Find the area of a semicircle whose diameter is 28cm​

The length of JK is 18.333.

Since MN is parallel to JK, the angles formed by JLN and MLK are equal. Therefore, we can use the Triangle Proportionality Theorem, which states that if a line parallel to one side of a triangle divides the other two sides proportionally, then the triangles are similar.

Using the Triangle Proportionality Theorem, we can set up the following proportion:

[tex]$\frac{LK}{JL} = \frac{MN}{LN}$[/tex]

Therefore,

[tex]$\frac{7.2}{13.2} = \frac{10}{6}$[/tex]

We can cross-multiply to solve for JK:

[tex]$7.2 \cdot 6 = 13.2 \cdot 10$\\$43.2 = 132$\\$JK = \frac{132}{7.2} = 18.333$[/tex]

Therefore, the length of JK is 18.333.

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Answer:

a) The middle 95 percent of all miniature Tootsie Rolls will fall between 3.04 grams and 3.56 grams.

b) 6.18% probability that a randomly chosen miniature Tootsie Roll will weigh more than 3.50 grams.

c) 52.29% probability that a randomly chosen miniature Tootsie Roll will weigh between 3.25 and 3.45 grams

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z=\dfrac{X-\mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The Empirical Rule is also used to solve this question. It states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

[tex]\mu=3.30,\sigma=0.13[/tex]

(a) Within what weight range will the middle 95 percent of all miniature Tootsie Rolls fall?

By the Empirical Rule the weight range of the middle 95% of all miniature Tootsie Rolls fall within two standard deviations of the mean. So

3.30 - 2 x 0.13 = 3.04

3.30 + 2 x 0.13 = 3.56

The middle 95 percent of all miniature Tootsie Rolls will fall between 3.04 grams and 3.56 grams.

(b) What is the probability that a randomly chosen miniature Tootsie Roll will weigh more than 3.50 grams?

This probability is 1 subtracted by the p-value of Z when X = 3.50. So

[tex]Z=\dfrac{X-\mu}{\sigma}[/tex]

[tex]Z=\dfrac{3.50-3.30}{0.13}[/tex]

[tex]Z=1.54[/tex]

[tex]Z=1.54[/tex] has a p-value of 0.9382.

1 - 0.9382 = 0.0618

6.18% probability that a randomly chosen miniature Tootsie Roll will weigh more than 3.50 grams.

c) What is the probability that a randomly chosen miniature Tootsie Roll will weigh between 3.25 and 3.45 grams?

This is the p-value of Z when X = 3.45 subtracted by the p-value of Z when X = 3.25. So

X = 3.45

[tex]Z=\dfrac{X-\mu}{\sigma}[/tex]

[tex]Z=\dfrac{3.45-3.30}{0.13}[/tex]

[tex]Z=1.15[/tex]

[tex]Z=1.15[/tex] has a p-value of 0.8749.

X = 3.25

[tex]Z=\dfrac{X-\mu}{\sigma}[/tex]

[tex]Z=\dfrac{3.25-3.30}{0.13}[/tex]

[tex]Z=-0.38[/tex]

[tex]Z=-0.38[/tex] has a p-value of 0.3520

0.8749 - 0.3520 = 0.5229

52.29% probability that a randomly chosen miniature Tootsie Roll will weigh between 3.25 and 3.45 grams.

Answer: The missing dimension of the rhombus in the given figure is Height of rhombus h  h=A/b=90/15= 6cm. so missing dimension is h=6cm

What is Dimension ?

In general, dimension refers to the measurement or size of an object, space, or quantity along a particular axis or direction. In mathematics, dimension refers to the number of coordinates needed to specify a point in a space.

What is Rhombus ?

A rhombus is a type of quadrilateral (a four-sided polygon) in which all four sides are of equal length. It is a special case of a parallelogram in which the opposite sides are parallel to each other, and its opposite angles are equal.

In the given question,

area of rhombus is A=b*h  so it can be rewritten as  h=A/b by substituting values given in question we get h= 6cm

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The dimensions of the rectangular frame is found as : 12 and 6 inches.

When a triangle is just a right triangle, the hypotenuse square is equal to the sum of the squares of the triangle's legs.

That's a picture frame, therefore pay attention that it must be rectangular.

Hence, the triangle is really a right triangle, and the Pythagorean theorem will eventually be applied.

You are aware that the square of the hypotenuse is 20 and equals 400.

hence, a²  + b²  = 400 and...

So because length is 4 times more than the breadth, a = b + 4.

This can be resolved if "b + 4" is substituted for "a":

(b + 4)² + b²  = 400,

(b + 4)(b + 4) + b²  = 400,

b² + 8b + 16 + b²  = 400,

2b² + 8b = 384

Further solving;

b² + 4b = 192

b² + 4b - 192 = 0

(b + 16)(b - 12) = 0

Due to the fact that a length cannot be negative, b must therefore be between b - 16 or 12 (negative value not taken)

The second leg is 12 + 4 = 6.

Thus,  the dimensions of the rectangular  frame is found as : 12 and 6 inches.

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Students were asked to simplify the expression using trigonometric identities:

 A.  student A is correct; student B was confused by the division

 B.  3: cos²(θ)/(sin(θ)csc(θ)); 4: cos²(θ)

Trigonometric Identities are equality statements that hold true for all values of the variables in the equation and that use trigonometry functions.

There are several distinctive trigonometric identities that relate a triangle's side length and angle. Only the right-angle triangle is consistent with the trigonometric identities.

The six trigonometric ratios serve as the foundation for all trigonometric identities. Sine, cosine, tangent, cosecant, secant, and cotangent are some of their names.

Each student correctly made use of the trigonometric identities

cosec(θ) = 1/sin(θ)

1 -sin²(θ) = cos²(θ)

A.

Student A's work is correct.

Student B apparently got confused by the two denominators in Step 2, and incorrectly replaced them with their quotient instead of their product.

The transition from Step 2 can look like:

[tex]\frac{(\frac{1-sin^2\theta}{sin\theta} )}{cosec\theta} =\frac{1-sin^2\theta}{sin\theta} .\frac{1}{cosec\theta} =\frac{cos^2\theta}{(sin\theta)(cosec\theta)}[/tex]

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Complete question:

Students were asked to simplify the expression the quantity cosecant theta minus sine theta end quantity over cosecant period Two students' work is given. (In image below)

Part A: Which student simplified the expression incorrectly? Explain the errors that were made or the formulas that were misused. (5 points)

Part B: Complete the student's solution correctly, beginning with the location of the error. (5 points)

Answer:

150

Step-by-step explanation:

Using the formulas

A=6a2

V=a3

Solving forA

A=6V⅔=6·125⅔ ≈150

Answer:

Step-by-step explanation:

If the volume of a cube is 125 cm³, it means that each side of the cube measures 5 cm (since 5 x 5 x 5 = 125).

To find the surface area of the cube, we need to calculate the area of each of the six faces and add them together.

The area of each face is simply the length of one side squared (or side x side).

So, the surface area of the cube would be:

6 x (5 cm x 5 cm) = 6 x 25 = 150 cm²

Therefore, the surface area of the cube is 150 cm².

Answer:

Let's denote the number of adult tickets sold by "x" and the number of children's tickets sold by "y".

We know that the total number of tickets sold is 960, so:

x + y = 960

We also know the total amount collected, which is $11,920:

15x + 11y = 11,920

Now we have two equations with two unknowns, and we can solve for x and y. One way to do this is by using substitution. We can solve the first equation for x:

x = 960 - y

Then substitute this expression for x into the second equation:

15(960 - y) + 11y = 11,920

Expanding the brackets gives:

14,400 - 15y + 11y = 11,920

Simplifying:

4y = 2,480

y = 620

Now we can use this value of y to find x:

x = 960 - y = 960 - 620 = 340

Therefore, 340 adult tickets and 620 children's tickets were sold.

Answer:

Let's use a system of equations to solve the problem.

Let x be the number of adult tickets sold and y be the number of children's tickets sold.

We know that the total number of tickets sold is 960, so we can write:

x + y = 960

We also know that the total revenue from the ticket sales was $11,920. The revenue from the adult tickets is $15 times the number of adult tickets sold, and the revenue from the children's tickets is $11 times the number of children's tickets sold. So we can write:

15x + 11y = 11,920

We now have two equations with two unknowns, which we can solve using substitution or elimination. Let's use elimination:

Multiply the first equation by 11 to get 11x + 11y = 10,560

Subtract the second equation from the first to get:

15x + 11y - 15x - 11y = 11,920 - 10,560

Simplifying, we get:

4x = 1,360

Dividing both sides by 4, we get:

x = 340

So 340 adult tickets were sold.

Substituting this value back into the first equation, we get:

340 + y = 960

Solving for y, we get:

y = 620

So 620 children's tickets were sold.

Therefore, there were 340 adult tickets sold and 620 children's tickets sold.

Answer:

A

Step-by-step explanation:

Congruent triangles

Answer:

A) IJ is congruent (equally long) to JG.

Step-by-step explanation:

KH splits IG and EF each into 2 equal halves.

the other answer options compare not-correlating distances, and so, they are not surprisingly not equally long.

Australian sheepdogs' life length follows a uniform distribution between 8 and 14 years.

a) The probability that a sheep dog will live at least 10 years, P( X ≥10) is equals to the 1/3

b) The probability that a sheepdog will live no more than 11 years, P( X ≤11) is equals to the 1/2.

c) The probability that a sheepdog will live between 10 and 13 years, P(10≤X≤13), is equals to the 1/2

In statistics, uniform distribution,is a distribution function where every possible result is equally likely, i.e., the probability of each event occurring is the equal. Here we have the length of Australian sheepdogs life follows a uniform distribution between 8 and 14 years. So, the probability distribution function, pdf for uniform distribution is f(x) = 1/(b - a), a<x<b , a= 8 , b = 14

=> f(x) = 1/(14 - 8) = 1/6

Area of uniform distribution= height × base and base = 14 - 8 = 6

height of uniform distribution is = 1/( Max - Min) = 1/6.

a) the probability that a sheepdog will live at least 10 years, P(X ≥10)= 1 - P(X< 10)

The cumulative distribution function in uniform distribution is P(X ≤ x) = (x − a)/(b − a) so, P( X < 10) = (10 - 8)/(14 - 8)

= 2/6 = 1/3

b) the probability that a sheepdog will live no more than 11 years, P( X ≤11)

= (11 - 8)/( 14 - 8)

= 3/6 = 1/2

c) In uniform distribution, P(c ≤ x ≤ d)

= (d-c)/(b- a)

The probability that a sheepdog will live between 10 and 13 years, P( 10≤X≤13), c = 10 , d = 13 , b = 14 , a = 8

=> [tex]P( 10≤X≤13) = \frac{d-c}{b- a} = \frac{13 - 10}{14 - 8} [/tex]

= 3/6 = 1/2

Hence required probability value is 1/2.

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Complete question:

Uniform Questions:

Australian sheepdogs have a relatively short life. The length of their life follows a uniform distribution between 8 and 14 years. Questions:

a)What is the probability that a sheepdog will live at least 10 years?

b)What is the probability that a sheepdog will live no more than 11 years?

c)What is the probability that a sheepdog will live between 10 and 13 years?

No, the graph of tan 0, -1r/2 < 0 < 1r/2, is not asymptotic to the lines x = i and x = -i.

An asymptote is a line that a graph approaches but never crosses. The graph of tan 0, -1r/2 < 0 < 1r/2, has a period of π, meaning it repeats after every π, and will never cross the lines x = i and x = -i. This can be seen in the equation y = tan 0, where the x-values of -1r/2 and 1r/2 are replaced with the x-values of i and -i. The equation would be y = tan(i) and y = tan(-i), and the graphs of these equations would not be asymptotic to the lines x = i and x = -i.No, the graph of tan 0, -1r/2 < 0 < 1r/2, is not asymptotic to the lines x = i and x = -i.

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The point estimate for the population mean age at first marriage is 21.89, the true population mean age at first marriage falls between 21.815 and 21.965 years with a small margin of error due to a large sample size. A 99% confidence interval is (21.836, 21.944).

The point estimate at first marriage is 21.89.

We can interpret the 99% confidence interval as follows: we are 99% confident that the true population mean age at first marriage falls between 21.815 and 21.965 years.

To compute a 95% confidence interval, we can use the formula:

Margin of error = z*(SE)

where z is the z-score corresponding to the desired confidence level (1.96 for 95% confidence), and SE is the standard error of the mean, which is equal to the standard deviation divided by the square root of the sample size.

Thus, for the given data:

Margin of error = 1.96*(4.787/sqrt(26920)) = 0.054

The 95% confidence interval can be computed as:

21.89 ± 0.054

which gives us a range of (21.836, 21.944).

The margin of error for this confidence interval is small because the sample size is very large (n=26920). As the sample size increases, the standard error of the mean decreases, which in turn reduces the margin of error.

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_____The given question is incomplete, the complete qustion is given below:

a. Use the summary to determine the point estimate of the population mean and margin of error for the confidence interval

b. interpret the confidence interval

c. verify the results by computing a 95% confidence interval with the information provided

d. why is the margin of error for this confidence interval so small? A study asked respondents, "If ever married, how old were you when you first married? The results are summarized in the technology excerpt that follows. Complete parts (a) through (d) below. One-Sample T: AGEWED Variable N Mean StDev SE Mean 99.0% CI AGEWED 26920 21.890 4.787 0.029 (21.815, 21.965) L attention and maintarhaan Hansen

(a) The country will first experience shortages of food in approximately 26.6 years

(b) If the country doubled its initial food supply and maintained a constant rate of increase in the supply, shortages would still occur in approximately 38 years.

(c) If the country doubled the rate at which its food supply increases, in addition to doubling its initial food supply, shortages would still occur in approximately 55.4 years.

a. Let P(t) be the population of the country at time t (in years), and F(t) be the food supply of the country at time t.

We know that P(0) = 4 million, and P'(t) = 0.05P(t), which means that the population is increasing by 5% per year.

We also know that F(0) = 8 million, and F'(t) = 0.25 million, which means that the food supply is increasing by 0.25 million people per year.

When the food supply is just enough to feed the population, we have P(t) = F(t), so we can solve for t as follows:

4 million x (1 + 0.05)^t = 8 million + 0.25 million x t

[tex]4(1 + 0.05)^t = 8 + 0.25t\\\\t \approx 26.6 \ years[/tex]

b. If the country doubled its initial food supply, then F(0) = 16 million. We can use the same equation as before and solve for t:

4 million x  (1 + 0.05)^t = 16 million + 0.25 million x t

[tex]4(1 + 0.05)^t = 16 + 0.25t\\\\t \approx 38 \ years[/tex]

c. If the country doubled the rate at which its food supply increases and doubled its initial food supply, then we have F(0) = 16 million and F'(t) = 0.5 million. Using the same equation as before, we get:

4 million x  (1 + 0.05)^t = 32 million + 0.5 million x t

[tex]4(1 + 0.05)^t = 32 + 0.5t\\\\t \approx 55.4 \ years[/tex]

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Answer:

1.066 (3 d.p.)

Step-by-step explanation:

The volume of the solid formed by revolving a region, R, around a vertical axis, bounded by x = a and x = b, is given by:

[tex]\displaystyle 2\pi \int^b_ar(x)h(x)\;\text{d}x[/tex]

where:

[tex]\hrulefill[/tex]

We want to find the volume of the solid formed by revolving a region, R, around the y-axis, where R is bounded by:

[tex]y=\dfrac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{3}}[/tex]

[tex]y=0[/tex]

[tex]x=0[/tex]

[tex]x=1[/tex]

As the axis of rotation is the y-axis, r(x) = x.

Therefore, in this case:

[tex]r(x)=x[/tex]

[tex]h(x)=\dfrac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{3}}[/tex]

[tex]a=0[/tex]

[tex]b=1[/tex]

Set up the integral:

[tex]\displaystyle 2\pi \int^{1}_0x \cdot\dfrac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{3}}\;\text{d}x[/tex]

Take out the constant:

[tex]\displaystyle 2\pi \cdot \dfrac{1}{\sqrt{2\pi}}\int^{1}_0x \cdot e^{-\frac{x^2}{3}}\;\text{d}x[/tex]

[tex]\displaystyle \sqrt{2\pi}\int^{1}_0x \cdot e^{-\frac{x^2}{3}}\;\text{d}x[/tex]

Integrate using the method of substitution.

[tex]\textsf{Let}\;u=-\dfrac{x^2}{3}\implies \dfrac{\text{d}u}{\text{d}x}=-\dfrac{2x}{3}\implies \text{d}x=-\dfrac{3}{2x}\;\text{d}u[/tex]

[tex]\textsf{When}\;x=0 \implies u=0[/tex]

[tex]\textsf{When}\;x=1 \implies u=-\dfrac{1}{3}[/tex]

Rewrite the original integral in terms of u and du:

[tex]\displaystyle \sqrt{2\pi}\int^{-\frac{1}{3}}_0x \cdot e^{u}\cdot -\dfrac{3}{2x}\;\text{d}u[/tex]

[tex]\displaystyle \sqrt{2\pi}\int^{-\frac{1}{3}}_0 -\dfrac{3}{2}e^{u}\; \text{d}u[/tex]

[tex]-\dfrac{3\sqrt{2\pi}}{2}\displaystyle \int^{-\frac{1}{3}}_0 e^{u}\; \text{d}u[/tex]

Evaluate:

[tex]\begin{aligned}-\dfrac{3\sqrt{2\pi}}{2}\displaystyle \int^{-\frac{1}{3}}_0 e^{u}\; \text{d}u&=-\dfrac{3\sqrt{2\pi}}{2}\left[ \vphantom{\dfrac12}e^u\right]^{-\frac{1}{3}}_0\\\\&=-\dfrac{3\sqrt{2\pi}}{2}\left[ \vphantom{\dfrac12}e^{-\frac{1}{3}}-e^0\right]\\\\&=-\dfrac{3\sqrt{2\pi}}{2}\left[ \vphantom{\dfrac12}e^{-\frac{1}{3}}-1\right]\\\\&=1.06582594...\\\\&=1.066\; \sf (3\;d.p.)\end{aligned}[/tex]

Therefore, the volume of the solid is approximately 1.066 (3 d.p.).

[tex]\hrulefill[/tex]

[tex]\boxed{\begin{minipage}{3 cm}\underline{Integrating $e^x$}\\\\$\displaystyle \int e^x\:\text{d}x=e^x(+\;\text{C})$\end{minipage}}[/tex]

The command that will allow you to set the interface link as preferred over another in OSPF is the "ip ospf cost 10" command. The correct answer is D).

The OSPF protocol uses the concept of cost to determine the best path for routing information to a remote network. The cost of a particular path is calculated based on the bandwidth of the interface and is inversely proportional to it. Therefore, the lower the cost, the more preferred the path is.

To set the preferred link, you can decrease the cost of the desired interface using the "ip ospf cost" command. For example, if you want to set the cost of interface GigabitEthernet0/0 to 10, you can use the following command:

ip ospf cost 10 interface GigabitEthernet0/0

This will make OSPF prefer the specified interface over other interfaces with a higher cost, ensuring that traffic is routed through the preferred link.

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____The given question is incomplete, the complete question is given below:

You need to setup a preferred link that OSPF will use to route information to a remote network. Which command will allow you to set the interface link as preferred over another?

A. ip ospf preferred 10

B. ip ospf priority 10

C. ospf bandwidth 10

D. ip ospf cost 10

Answer:

If the wholesale price of the king-size mattress is $359.00, and the specialty store marks up the price by 40%, the price Debra would pay at the specialty store is:

Wholesale price + Mark-up amount = Price at specialty store

$359.00 + 40% of $359.00 = $359.00 + $143.60 = $502.60

Therefore, Debra would pay $502.60 for the mattress at the specialty store.

To approximate the root of √5 with a precision of 0.0078125, the binary search method must be used. The process is repeated until the desired value is reached.

The Intermediate Value Theorem states that if a continuous function takes on two different values at two different points, then it must take on a value in between those two points.

By using the function f(x) = x²-5, it can be seen that f(2) < 0 and f(3) > 0. This means that there must be a value, 2 < c < 3, such that f(c) = 0.

The next step is to determine if f(2.5) is less than or greater than 0. If it is the same sign as 2.5, the endpoint of the same sign must be used as the new endpoint of the interval.

By using the Intermediate Value Theorem, it was proven that there must be a value between 2 and 3 that satisfies the equation f(x) = x²-5. The midpoint of those two points was used as the first approximation, and by repeating the process of binary search, the desired precision was achieved.

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Answer:

the answer to that is -31

To satisfy the inequality x less-than StartFraction 7 over 12 EndFraction.

Inequalities are called as the mathematical expressions in which both sides are nonequal. Unlike to equations, we compare two values in inequality. Less than (or less than or equal to), greater than (or greater than or equal to), or not equal to signs can be used in place of the equal sign in between.

The inequality is 1/4 + x < 5/6 in order to solve this inequality we need to isolate the value of x, that is our variable of interest. This is shown bellow:

1/4 + x < 5/6

x < 5/6 - 1/4

LMC is used to subtract the fractions we have as follows:

x < (2*5 - 3*1)/12

x < (10 - 3)/12

x< 7/12

The inequality must be satisfied for x to be smaller than 7/12.

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Answer:  x < 7/12

Step-by-step explanation:

The answer to this question according to given information is 10. The intersection of the sets A and C, denoted as (A∩C), is the set of all elements present in both A and C.

Intersection is a term used to describe the common elements between two or more sets. A set is called as a collection of distinct objects, usually represented by a list, table, or diagram.  In other words, it is the overlap between the two sets. For example, the intersection of the set of even numbers and the set of prime numbers would be the set of all even prime numbers.

This is calculated by finding the common elements in the two sets. In this case, (A∩C) = {10, 13, 17, 23}.

Next, the intersection of (A∩C) and B, denoted as (A∩C)∩B, is the set of all elements which are common to all three sets. This is calculated by finding the common elements in (A∩C) and B. In this case, (A∩C)∩B = {10, 17}.

Finally, the number of elements in the set (A∩C)∩B is 10, since there are 10 elements in the set. To summarize, the number of elements in the set (A∩C)∩B is 10.

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The number of elements in the set (A∩C)∩B is 10. The intersection of the sets A and C, denoted as (A∩C), is the set of all elements present in both A and C.

Intersection is a term used to describe the common elements between two or more sets. A set is called as a collection of distinct objects, usually represented by a list, table, or diagram.  

This is calculated by finding the common elements in the two sets.

In this case, (A∩C) = {10, 13, 17, 23}.

Next, the intersection of (A∩C) and B, denoted as (A∩C)∩B, is the set of all elements which are common to all three sets.

This can be calculated by finding the common elements in (A∩C) and B.

In this case, (A∩C)∩B = {10, 17}.

Finally, the number of elements in the set (A∩C)∩B is 10, since there are 10 elements in the set.

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The given statement " let v be a vector space. we know that v must contain a zero vector, 0v and the zero vector is unique." is true and proved as an element in vector satisfies the define of zero vector i.e.,

0v + u  = u.

To show that the zero vector is unique, we need to prove that there can be only one element in the vector space that satisfies the definition of a zero vector, namely:

For any vector u in v, 0v + u = u + 0v = u.

To do this, suppose that there exist two distinct zero vectors, 0v and 0'v, such that 0v ≠ 0'v. Then, by the definition of a zero vector, we have:

0v + 0'v = 0'v + 0v = 0'v.

But, by the associative property of vector addition, we can also write:

0v + 0'v = (0v + u) + (-u + 0'v) = u + (-u) = 0v.

Similarly, we can write:

0'v + 0v = (0'v + u) + (-u + 0v) = u + (-u) = 0'v.

These equations show that 0v = 0'v, which contradicts our assumption that 0v ≠ 0'v. Therefore, the zero vector is unique, and there can be only one element in the vector space that satisfies the definition of a zero vector.

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Answer:

[tex]p(x)=(x+5)(x-3)(x+4)[/tex]

Step-by-step explanation:

Given : [tex]p(x)=x^3+6x^2-7x-60[/tex]

Solution :

Part A:

First find the potential roots of p(x) using rational root theorem;

So, [tex]\text{Possible roots = }\pm\frac{\text{factors of constant term}}{\text{factors of leading coefficient}}[/tex]

Since constant term = -60

Leading coefficient = 1

[tex]\text{Possible roots = }\pm\frac{\text{factors of 60}}{\text{factors of 1}}[/tex]

[tex]\text{Possible roots = }\pm\frac{\text{1,2,3,4,5,6,10,12,15,20,60}}{\text{1}}[/tex]

Thus the possible roots are [tex]\pm1,\pm2,\pm3,\pm4,\pm5,\pm6,\pm10,\pm12,\pm15,\pm20,\pm60[/tex]

Thus from the given options the correct answers are -10, -5, 3, 15

Now For Part B we will use synthetic division

Out of the possible roots we will use the root which gives remainder 0 in synthetic division :

Since we can see in the figure With -5 we are getting 0 remainder.

Refer the attached figure

We have completed the table and have obtained the following resulting coefficients: 1 , 1,−12,0.  All the coefficients except the last one are the coefficients of the quotient, the last coefficient is the remainder.

Thus the quotient is

And remainder is 0 .

So to get the other two factors of the given polynomial we will solve the quotient by middle term splitting

[tex]x^2+x-12=0[/tex]

[tex]x^2+4x-3x-12=0[/tex]

[tex]x(x+4)-3(x+4)=0[/tex]

[tex](x-3)(x+4)=0[/tex]

Thus x - 3 and x + 4 are the other two factors

So, p(x)=(x+5)(x-3)(x+4)

Answer:

The slope of a line passing through two points (x1, y1) and (x2, y2) is given by the formula:

m = (y2 - y1) / (x2 - x1)

Substituting the values, we get:

m = (5 - (-10)) / (6 - 3) = 15/3 = 5

Now that we have the slope, we can use the point-slope form of a linear equation to write the equation of the line:

y - y1 = m(x - x1)

Substituting the values of m, x1, and y1, we get:

y - (-10) = 5(x - 3)

Simplifying and rearranging the equation, we get:

y + 10 = 5x - 15

y = 5x - 25

Therefore, the equation of the line passing through (3,-10) and (6,5) in slope-intercept form is y = 5x - 25.

Step-by-step explanation:

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Using the two points you have plotted, draw the parabola. It should look like a downward-facing curve opening at the vertex (0, 7).

What is parabola?

A parabola is a symmetrical, U-shaped curve that is formed by the graph of a quadratic function.

Assuming you meant [tex]f(x) = -x^2 + 7[/tex], here's how you can graph the parabola using the parabola tool:

Find the vertex

The vertex of the parabola is located at the point (-b/2a, f(-b/2a)), where a is the coefficient of the [tex]x^2[/tex] term and b is the coefficient of the x term. In this case, a = -1 and b = 0, so the vertex is located at the point (0, 7).

Plot the vertex

Using the parabola tool, plot the vertex at the point (0, 7).

Plot a second point

To plot a second point, you can choose any x value and find the corresponding y value using the quadratic function. For example, if you choose x = 2, then [tex]f(2) = -2^2 + 7 = 3[/tex]. So the second point is located at (2, 3).

Therefore, Using the two points you have plotted, draw the parabola. It should look like a downward-facing curve opening at the vertex (0, 7).

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Complete Question:

Use the parabola tool to graph the quadratic function.

f(x) = -√² +7

Graph the parabola by first plotting its vertex and then plotting a second point on the parabola.

Answer:

8

Step-by-step explanation:

Each side of the. decagon is equal so each side is 8

The derivative of f(x) is

[tex]f'(x) = -5pi x^(pi-1) + 6.1 * 5.1x^(5.1-1) + 5.1pi^(5.1-1)[/tex].

The derivative of a function is a measure of how that function changes as its input changes. Derivatives are also used in calculus to find the area under a curve, or to solve differential equations.

In this case, the function f(x) is a polynomial, which means it is a combination of terms of the form [tex]ax^b[/tex], where a and b are constants. The derivative of f(x) can be calculated by taking the derivative of each term in the function and then combining them together.

The derivative of a term [tex]ax^b[/tex] is [tex]abx^(b-1)[/tex]. For the first term of f(x),[tex]-5x^pi[/tex], the derivative is [tex]-5pi x^(pi-1)[/tex]. For the second term, [tex]6.1x^5.1[/tex] the derivative is[tex]6.1 * 5.1x^(5.1-1)[/tex]. For the third term, [tex]pi^5.1[/tex], the derivative is [tex]5.1pi^(5.1-1)[/tex].

Combining these terms together, the derivative of f(x) is

[tex]f'(x) = -5pi x^(pi-1) + 6.1 * 5.1x^(5.1-1) + 5.1pi^(5.1-1)[/tex].

This answer is the derivative of the given function. This is how the function changes as its input changes.

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The derivative of f(x)= [tex]-5x^{\pi}+6.1x^{5.1}+\pi^{5.1}[/tex] is  [tex]-5\pi x^{\pi -1}[/tex]+  [tex]6.1*5.1x^{5.1-1}[/tex] +5.1[tex]\pi^{5.1-1}[/tex] which can be calculated with the power rule.

The derivative of a function is a measure of how that function changes as its input changes. Derivatives are also used in calculus to find the area under a curve, or to solve differential equations.

The derivative of the given function f(x) = [tex]-5x^{\pi}+6.1x^{5.1}+\pi^{5.1}[/tex] can be calculated with the power rule, which states that the derivative of xⁿ is nx⁽ⁿ⁻¹⁾

To calculate the derivative of the given function, we begin by applying the power rule to each term.

The first term is [tex]-5^{\pi }[/tex] which has a derivative of [tex]-5\pi x^{\pi -1}[/tex].

The second term is [tex]6.1x^{5.1}[/tex] which has a derivative of [tex]6.1*5.1x^{5.1-1}[/tex].

The third term is [tex]\pi^{5.1}[/tex], which has a derivative of 5.1[tex]\pi^{5.1-1}[/tex].

Therefore, the derivative of the given function

f(x)= [tex]-5x^{\pi}+6.1x^{5.1}+\pi^{5.1}[/tex] is  [tex]-5\pi x^{\pi -1}[/tex]+  [tex]6.1*5.1x^{5.1-1}[/tex] +5.1[tex]\pi^{5.1-1}[/tex].

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Question:

Compute the derivative of the given function.

f(x) = - [tex]5x^{\pi }[/tex]+[tex]6.1x^{5.1}[/tex]+[tex]\pi^{5.1}[/tex]

​

The integral expression for the volume of the solid formed by revolving the region bounded by the graphs of y = x₂ - 3x and y = x about the horizontal line y = 6 is  2πx(6 - x² + 3x)dx, which is integrated from x=0 to x=3, which gives us   81π/2.

To find the integral expression for the volume of the solid formed by revolving the region bounded by the graphs of y=x² - 3x and y=x about the horizontal line y=6, we can use the method of cylindrical shells.

First, we need to find the limits of integration, The graphs of y = x² - 3x  and y=x intersect at x=0 and x=3. Therefore, we integrate from x=0 to x=3.

Next, we consider a vertical strip of width dx at a distance x from the y- boxes. the height of the strip is the difference between the height of the curve y=  x² - 3x and the line y=6, which is   6 - (x² - 3x) = 6 - x² + 3x. the circumference of the shell is 2π times the distance x from the y-axis, and the thickness of the shell is dx. the volume of the shell is the product of the height, circumference, and thickness which is

dV = 2πx(6 - x² + 3x)dx

To find the total volume, we integrate this expression from x=0 to x=3.

V = ∫₀³ 2πx(6 - x² + 3x)dx, after simplifying the integrand we get :

V = 2π ∫₀³ (6x - x³ + 3x²)dx, integrating term by term we get :

V = 2π [(3x²/2) - (x⁴/4) + (x^3)] from 0 to 3, now  evaluation at the limits of integration we get:

V = 2π [(3(3)²/2) - ((3)⁴/4) + (3)³] - 2π [(0)^2/2 - ((0)⁴/4) + (0)^3]=  2π [(27/2) - (27/4) + 27] - 0 = 81π/2

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The standard form of the equation of the ellipse is (x - 2)^2 / 4 + (y - 3)^2 / 7 = 1

To find the standard form of the equation of the ellipse, we first need to determine some of its properties.

The foci of the ellipse are given as (2, 0) and (2, 6). This tells us that the center of the ellipse is at the point (2, 3), which is the midpoint of the line segment connecting the foci.

The major axis of the ellipse is given as a length of 8. Since the major axis is the longest dimension of the ellipse, we can assume that the length of the major axis is 2a = 8, so a = 4.

Next, we need to determine the length of the minor axis. We know that the distance between the foci is 2c = 6, so c = 3. Since c is the distance from the center of the ellipse to each focus, we can use the Pythagorean theorem to find the length of the minor axis

b^2 = a^2 - c^2

b^2 = 4^2 - 3^2

b^2 = 7

b = sqrt(7)

Now we have all the information we need to write the standard form of the equation of the ellipse. The standard form is

(x - h)^2 / a^2 + (y - k)^2 / b^2 = 1

where (h, k) is the center of the ellipse. Plugging in the values we found, we get

(x - 2)^2 / 4 + (y - 3)^2 / 7 = 1

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Step-by-step explanation:

The formula for calculating the balance on a CD (Certificate of Deposit) after a certain amount of time is:

A = P(1 + r/n)^(nt)

Where: A = the ending balance P = the principal (initial investment) r = the annual interest rate (as a decimal) n = the number of times interest is compounded per year t = the time in years

In this case, the initial investment is $1,800.00, the annual interest rate is 2.3% (or 0.023 as a decimal), and the investment period is 2 years. Assuming that the interest is compounded annually, we can substitute these values into the formula:

A = 1800(1 + 0.023/1)^(1*2) A = 1800(1.046729) A = 1883.12

Rounding to the nearest cent, the ending balance after 2 years on the CD is $1,883.75 (option B). Therefore, option B is the correct answer.

His fortnightly income is $3000 where the yearly salary is $78000 using 26 fortnights in a year.

Fortnightly income is the amount of income a person earns every two weeks. It is usually calculated by dividing the person's yearly income by the number of fortnights in a year, which is typically 26.

To calculate the fortnightly income (F), we need to divide the yearly salary (Y) by the number of fortnights in a year (N), which is 26.

So mathematically, we can express the calculation of the fortnightly income as:

F = Y / N

Substituting the given values, we get:

F = $78000 / 26

Simplifying the expression, we get:

F = $3000

Therefore, his fortnightly income is $3000.

For example, if a person's yearly salary is $52,000, their fortnightly income would be calculated as:

Fortnightly income = Yearly salary / Number of fortnights

Fortnightly income = $52,000 / 26

Fortnightly income = $2,000

The person's fortnightly income would be $2,000.

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The complete question is Tom’s yearly salary is $78000. Calculate Tom’s fortnightly income. (Use 26 fortnights in a year). Fortnightly income = ?$