Simplify to an expression involving a single trigonometric function with no fractions.


cos(−x)+tan(−x)sin(−x)

True. If a counting number with two or more digits remains the same with its digits reversed, then the counting number is a multiple of 11.

When a two-digit number is reversed, it becomes a new number with the digits swapped, e.g., 12 becomes 21. The difference between the original number and the reversed number is obtained by subtracting one from the other. For example, the difference between 12 and 21 is 9. It can be observed that the difference between any two-digit number and its reverse is always a multiple of 9.

Now, let's consider the three-digit number ABC. When this number is reversed, it becomes CBA. The difference between the two is

which is a multiple of 11.

Therefore, if a counting number with two or more digits remains the same with its digits reversed, then the counting number is a multiple of 11.

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Complete Question:

If a counting number with two or more digits remains the same with its digits reversed, then the counting number is a multiple of 11. True/ False.

The answer of the given question based on the changing the denominator of fraction the answer is the fraction a+3/6-2a can be rewritten with a denominator of 2(a²-9) as (3 + a)/(2(a - 3)).

In mathematics,  formula is  mathematical expression or equation that describes  relationship between two or more variables or quantities. A formula can be used to solve problems or make predictions about particular situation or set of data.

Formulas often involve mathematical symbols and operations, like addition, subtraction, multiplication, division, exponents, and square roots. They may also include variables, which are typically represented by letters, and constants, which are fixed values that do not change.

To change the denominator of the fraction a+3/6-2a to 2(a²-9), we need to factor the denominator of the original fraction and then use algebraic manipulation to rewrite it in the desired form.

First, we can factor the denominator of the original fraction as follows:

6 - 2a = 2(3 - a)

Next, we can rewrite the denominator using the difference of squares formula:

2(a² - 9) = 2(a + 3)(a - 3)

Now, we can use the factored form of the denominator to rewrite the original fraction:

(a + 3)/(6 - 2a) = (a + 3)/(2(3 - a)) = -(a + 3)/(-2(a - 3)) = (3 + a)/(2(a - 3))

Therefore, the fraction a+3/6-2a can be rewritten with a denominator of 2(a²-9) as (3 + a)/(2(a - 3)).

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The score at the 20th percentile is 27.

To find the score at the 20th percentile of the 29 ages for Academy Award winning best actors, follow the steps below:

Arrange the given ages from smallest to largest.

18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77

Determine the total number of data points

n = 29

Find the rank of the percentile

20th percentile = (20/100) * 29 = 5.8 = 6 (rounded to the nearest whole number).The rank of the percentile is 6.

Use the rank to determine the corresponding data value. The corresponding data value is the value at the 6th position when the data is arranged in ascending order. The score at the 20th percentile is 27.

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The best conclusion from this hypothesis test in the context of the problem is that we can reject the null hypothesis. The null hypothesis for this problem is that the errors is not greater than the population mean.

The null hypothesis for this problem is that the population mean number of errors for the ethanol group is not greater than the population mean number of errors for the placebo group.

In other words, the null hypothesis is: H₀: μe ≤ μp. The alternative hypothesis is that the population mean number of errors for the ethanol group is greater than the population mean number of errors for the placebo group. In other words, the alternative hypothesis is: H₁: μe > μp.

The p-value is the probability of getting a test statistic at least as extreme as the one observed, assuming the null hypothesis is true. In this case, the p-value is 0.0106, which is less than the significance level of 0.01. This means that the observed test statistic is significant at the 1% level, and we reject the null hypothesis.

Therefore, we conclude that there is evidence to suggest that the population mean number of errors for the ethanol group is greater than the population mean number of errors for the placebo group.

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With respect to the average cost curves, the marginal cost curve: intersects both average total cost and average variable cost at their minimum points that is option B.

The fixed cost per unit of production is the average fixed cost (AFC). AFC will reduce consistently as output grows since total fixed costs stay constant. The variable cost per unit of production is known as the average variable cost (AVC). AVC generally declines until it reaches a minimum and then increases due to the growing and then lowering marginal returns to the variable input. The average total cost curve's (ATC) behaviour is determined by the behaviour of the AFC and AVC.

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Complete question:

With respect to the average cost curves, the marginal cost curve:

A) Intersects average total cost, average fixed cost, and average variable cost at their minimum point

B) Intersects both average total cost and average variable cost at their minimum points

C) Intersects average total cost where it is increasing and average variable cost where it is decreasing

D) Intersects only average total cost at its minimum point

[tex]\qquad \qquad \textit{inverse proportional variation} \\\\ \textit{\underline{y} varies inversely with \underline{x}} ~\hspace{6em} \stackrel{\textit{constant of variation}}{y=\cfrac{\stackrel{\downarrow }{k}}{x}~\hfill } \\\\ \textit{\underline{x} varies inversely with }\underline{z^5} ~\hspace{5.5em} \stackrel{\textit{constant of variation}}{x=\cfrac{\stackrel{\downarrow }{k}}{z^5}~\hfill } \\\\[-0.35em] ~\dotfill[/tex]

[tex]\stackrel{\textit{"y" varies inversely with "x"}}{y = \cfrac{k}{x}}\hspace{5em}\textit{we also know that} \begin{cases} x=25\\ y=100 \end{cases} \\\\\\ 100=\cfrac{k}{25}\implies 2500=k\hspace{12em}\boxed{y=\cfrac{2500}{x}} \\\\\\ \textit{when x = 10, what's "y"?}\qquad y=\cfrac{2500}{10}\implies y=250[/tex]

From the given graph, CE is longer than CD.

The length of the line segment bridging two locations in a plane is known as the distance between the points. d=√((x₂ - x₁)²+ (y₂ - y₁)²) is a common formula to calculate the distance between two points. This equation can be used to calculate the separation between any two locations on an x-y plane or coordinate plane.

Coordinates of E(8,6)

Coordinates of C(6,1)

Coordinates of D(3,-3)

x=8, y=6

x=6, y=1

x=3, y=-3

Distance CE=√{(8-6)² +(6-1)²} = √29

Distance CD=√{(6-3)² +(1+3)²}= √25=5

Therefore, CE is longer than CD.

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None of the available options match the parabola's equation.

To determine the equation of a parabola, we can utilize the vertex form. Assuming we can read the coordinates (h,k) from the graph, the aim is to utilize the coordinates of its vertex (maximum point, or minimum point), to formulate its equation in the form y=a(xh)2+k, and then to determine the value of the coefficient a.

A parabola's vertex form is given by:

[tex]y = a(x-h)^2 + k[/tex]

where (h,k) is the parabola's vertex.

[tex]y = a(x-5)^2 - 3[/tex]

These values are substituted into the equation to produce:

[tex]-15 = a(2-5)^2 - 3[/tex]

[tex]-15 = 9a - 3[/tex]

[tex]-12 = 9a[/tex]

[tex]a = -4/3[/tex]

[tex]y = (-4/3)(x-5)^2 - 3[/tex]

This equation is expanded and simplified to produce:

[tex]y = (-4/3)x^2 + (32/3)x - 53[/tex]

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In the following question, among the conditions given, The volume of the solid generated by revolving the plane region about the x-axis is (128/3)π.

To find the volume of the solid generated by revolving the plane region about the x-axis, we can use the shell method. The given region is bounded by the lines x=2, y=1 and x+y^2=4.

The integral to evaluate is:
V = 2π ∫r2h dx,
where r = x+y^2 = 4, h = y = 1, and x varies from 2 to 4.

Therefore, V = 2π ∫4^2*1 dx, from x = 2 to x = 4.

Evaluating the integral, we have:
V = 2π[4x^3/3]24
V = 2π(64/3 - 8/3)
V = (128/3)π
Therefore, the volume of the solid generated by revolving the plane region about the x-axis is (128/3)π.

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Answer:

This is not my own answer it is a copied one.

If h (x) represents the amount of money spent and x the amount of friends, then we can write it as in a pair as (x, h (x)) Then the pair given is (6.5, $92.25) Here you see a problem, x is 6.5, knowing that x represents the amount of friends, this is a problem because you need to have a whole number ( you can't have a 0.5 of a friend)

The estimated value of the machine after 4 years when the rate of depreciation dV/dt is inversely proportional to the square of t + 1 is $234,375.

Since the rate of depreciation is inversely proportional to the square of t + 1, we can write:

dV/dt = k / (t + 1)²

where k is the constant of proportionality. We can find k by using the initial value of the machine:

dV/dt = k / (t + 1)² = -100,000 / year when t = 0 (the first year)

Therefore, k = -100,000 * (1²) = -100,000.

To find the value of the machine after 4 years, we need to solve the differential equation:

dV/dt = -100,000 / (t + 1)

We can do this by separating variables and integrating:

∫dV / (V - 500,000) = ∫-100,000 dt / (t + 1)²

ln|V - 500,000| = 100,000 / (t + 1) + C

where C is the constant of integration.

We can find C by using the initial value of the machine:

ln|500,000 - 500,000| = 0 = 100,000 / (0 + 1) + C

Therefore, C = -100,000.

Substituting this value of C, we get:

ln|V - 500,000| = 100,000 / (t + 1) - 100,000

ln|V - 500,000| = -100,000 / (t + 1) + ln|e¹⁰|

ln|V - 500,000| = ln|e¹⁰ / (t + 1)²|

V - 500,000 = [tex]e^{10/(t + 1)²)}[/tex]

V = [tex]e^{10/(t + 1)²)}[/tex] + 500,000

Finally, we can estimate the value of the machine after 4 years by substituting t = 3:

V = [tex]e^{10/(3 + 1)²}[/tex] + 500,000

V ≈ $234,375

Therefore, the correct answer is $234,375.

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The expression for the marginal revenue function is MR(q) = 600r - 7.

The given product's revenue function is R(q) = - 7q + 600qr.

To find an expression for the marginal revenue function, we can use the following steps:

Step 1: Take the first derivative of the revenue function with respect to q to obtain the marginal revenue function MR(q).

Step 2: Simplify the expression for MR(q) to record the final result.

In other words, the marginal revenue function MR(q) is the derivative of the revenue function R(q) with respect to q. Here, R(q) = - 7q + 600qr.

So, we have to differentiate R(q) with respect to q to get MR(q).

The derivative of - 7q with respect to q is - 7.

The derivative of 600qr with respect to q is 600r because the derivative of q with respect to q is 1.

MR(q) = dR(q) / dq

= (d/dq)(- 7q + 600r)

= (- 7) + (600r)

= 600r - 7

The equation that represents the marginal revenue function is MR(q) = 600r - 7.

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Step 1: x is a positive number, so the absolute value of x will be equal to x.

Step 2: The expression x+|x| simplifies to 2x

Step 3: Therefore, the expression x+|x| = 2x if x≥0

The probability that at least 10 cabs pass you before you find one that is free is 0.00528665 or approximately 0.53%.

The solution to the problem is explained below:

Let, P(passes someone) = 0.15 or 15%

P(available taxi cab) = 0.85 or 85%

Let X be the number of cabs that pass before you find an available taxi cab. In order to find the probability that you see at least 10 cabs pass before you find a free one, we have to use the cumulative distribution function (CDF).

The probability that X is greater than or equal to 10 is equivalent to 1 - (the probability that X is less than 10). That is,P(X >= 10) = 1 - P(X < 10)

The probability that X is less than 10 is the probability of seeing 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 taxis pass you by.

Hence,P(X < 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)P(X = 0) = P(find an available taxi cab on the 1st attempt) = P(available taxi cab) = 0.85

P(X = 1) = P(find an available taxi cab on the 2nd attempt) = P(passed by the 1st taxi cab) x P(available taxi cab on the 2nd attempt) = (1 - P(available taxi cab)) x P(available taxi cab) = 0.15 x 0.85 = 0.1275

P(X = 2) = P(passed by the 1st taxi cab) x P(passed by the 2nd taxi cab) x P(available taxi cab on the 3rd attempt) = (1 - P(available taxi cab))² x P(available taxi cab) = 0.15² x 0.85 = 0.01817

P(X = 3) = (1 - P(available taxi cab))³ x P(available taxi cab) = 0.15³ x 0.85 = 0.002585

P(X = 4) = (1 - P(available taxi cab))⁴ x P(available taxi cab) = 0.15⁴ x 0.85 = 0.0003704

P(X = 5) = (1 - P(available taxi cab))⁵ x P(available taxi cab) = 0.15⁵ x 0.85 = 0.00005287

P(X = 6) = (1 - P(available taxi cab))⁶ x P(available taxi cab) = 0.15⁶ x 0.85 = 0.000007550

P(X = 7) = (1 - P(available taxi cab))⁷ x P(available taxi cab) = 0.15⁷ x 0.85 = 0.0000010825

P(X = 8) = (1 - P(available taxi cab))⁸ x P(available taxi cab) = 0.15⁸ x 0.85 = 0.000000154

P(X = 9) = (1 - P(available taxi cab))⁹ x P(available taxi cab) = 0.15⁹ x 0.85 = 0.0000000221

Hence,P(X < 10) = 0.85 + 0.1275 + 0.01817 + 0.002585 + 0.0003704 + 0.00005287 + 0.000007550 + 0.0000010825 + 0.000000154 + 0.0000000221 = 0.99471335

P(X >= 10) = 1 - P(X < 10) = 1 - 0.99471335 = 0.00528665

Therefore, the probability that at least 10 cabs pass you before you find one that is free is 0.00528665 or approximately 0.53%.

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The answer to the distributive property-based problem -3(-4x + 5) = [?]x - [] is 12x - 15.

A mathematical assertion proving the equality of two expressions is known as an equation. Variables, constants, and mathematical processes like addition, subtraction, multiplication, and division are frequently included. Finding the value of the variable that makes an equation correct is the aim of equation solving. Linear equations, quadratic equations, and exponential equations are just a few of the different ways that equations can be expressed.

given

-3(-4x + 5) = [?]x - [ ]

12x - 15

by distributive property

The answer to the distributive property-based problem -3(-4x + 5) = [?]x - [] is 12x - 15.

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The amount of interest earned on $4,000.00 for 4 years, compounding daily at 4.5% APR, is $1,064.08.

To calculate the amount of interest on $4,000.00 for 4 years, compounding daily at 4.5% APR, we can use the formula for compound interest:

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

where A is the final amount, P is the principal, r is the annual interest rate as a decimal, n is the number of times the interest is compounded per year, and t is the time in years.

In this case, we have P = $4,000.00, r = 0.045, n = 365 (since interest is compounded daily), and t = 4. Plugging these values into the formula, we get:

A = $4,000.00(1 + 0.045/365)^(365*4)

A = $4,000.00(1.0001234)^1460

A = $4,889.68

The final amount is $4,889.68, which means that the interest earned is:

Interest = $4,889.68 - $4,000.00 = $889.68

We are given that the monthly interest table shows that $1.197204 in interest is earned for each $1.00 invested. Therefore, to find the interest earned on $4,000.00, we can multiply the interest earned by the factor:

$1.197204 / $1.00 = 1.197204

Interest earned = $889.68 x 1.197204 = $1,064.08

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0.6990 is value of logarithm .

A logarithm is defined simply.

The logarithm represents the power to which a number must be raised to obtain another number (see Section 3 of this Math Review for more about exponents).

                              As an illustration, the base ten logarithm of 100 is 2, since ten multiplied by two equals 100: log 100 = 2, since 102 = 100. Binary logarithms, which have a base of 2, natural logarithms, which have a base of e 2.71828, and common logarithms with a base of 10 are the four most popular varieties of logarithms.

Now, log0.1= log(1/10) =log (10^-1) =-1 log10

Here log 10= 1 .

 log a = 0.05

log (100a) =  log (100 * 0.05)

                 =  log( 5.00)

                 = log(5)

                = 0.6990

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The work required to pump all the water out of the rectangular swimming pool over the top is approximately 2,323,200 ft-lb.

We have,

To find the work required to pump all the water out of the rectangular swimming pool, we can use the concept of work as the force multiplied by the distance.

First, let's calculate the weight of the water in the pool.

The weight of an object is given by the formula:

Weight = mass x gravitational acceleration

Since the density of water is given as 1 lb/ft³, we need to find the volume of water in the pool.

The volume of the pool is given by the formula:

Volume = length x width x depth

Volume = 50 ft x 30 ft x 6 ft = 9000 ft³

Now, let's calculate the weight of the water:

Weight = density x volume x gravitational acceleration

Weight = 1 lb/ft³ x 9000 ft³ x 32.2 ft/s² ≈ 290,400 lb

To pump all the water out over the top, we need to raise it to the height of the pool, which is 8 ft.

The work required to pump the water out is given by the formula:

Work = weight x height

Work = 290,400 lb x 8 ft = 2,323,200 ft-lb

Therefore,

The work required to pump all the water out of the rectangular swimming pool over the top is approximately 2,323,200 ft-lb.

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The cost of one small crate of apples is $12 and the cost of one large crate of peaches is $8. The cost of 7 small cates of apples and 5 large containers of peach is $126.

The system of equations that describe the question is:

3s + 10l = 116 equation 1

11s + 20l = 292 equation 2

Where:

s = cost of one small crate of apples

l = cost of one large crate of peaches

The elimination method would be used to determine the values of s and l.

Multiply equation 1 by 2

6s + 20l = 232 equation 3

Subtract equation 3 from equation 2:

5s = 60

Divide both sides of the equation by 5

s = 60 / 5

s = $12

Substitute for s in equation 1:

3(12) + 10l = 116

36 + 10l = 116

10l = 116 - 36

10l = 80

l = 80 / 10

l = 8

Cost of 7 small crates of apples and 5 large containers of peaches = (7 x 12) + (8 x 5) = $124

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Answer:

x = 1

Step-by-step explanation:

x + x + x + 30 = 33

3x + 30 = 33

3x + 30 - 30 = 33 - 30

3x = 3

x = 3/3 = 1

Answer:

4m + 6

Step-by-step explanation:

Since the length is twice as long your equation should look like this

2(2m + 3) = L

which would be 4m + 6 as the length of the rectangle

The velocity of a particle. P. moving along the x-axis is given by the differentiable function v, where (t) is measured is given by:

A differential function v gives the velocity of a particle P travelling down the x-axis, where v(t) is measured in metres per hour and t is measured in hours. v(t) is a declining function that is continuous. The table below shows several examples of v(t) values.

T [hours]                   0           2      4           7      10

v(t) [meters/hour] 20.3 14.4     10  7.3       5

a) We know that the particle's displacement is the area under the curve v(t). We can calculate the particle's displacement by integrating v(t). Because v(t) is a monotonous (constantly declining) differentiable function, it is also Riemann Integrable. There are now five non-uniform subdivisions:

Partition                     t0   t1 t2 t3 t4

T [hours]                      0           2 4 7 10

v(t) [meters/hour]   20.3 14.4 10 7.3 5

Using Right Riemann sum to approximate the displacement of particle between 0 hr and 10 hr is given by:

[tex]\sum_{n=1}^{4}v(t_n)\Delta t_n=v(t_1)(t_1-t_0)+v(t_2)(t_2-t_1)+v(t_3)(t_3-t_2)+v(t_4)(t_4-t_3) \\=(14.4)(2)+(10)(2)+(7.3)(3)+(5)(3) \\=28.8+20+21.9+15 \\=85.7[/tex]

Therefore, the total displacement between 0 hr and 10 hr is is 85.7 meters.

The estimated position of particle P at time t = 10 hour is 115.7 (= 30 +85.7) meters.

b) Because the function v(t) is decreasing and we are estimating the integral using the Right Riemann sum, the approximation in part(a) underestimates the displacement.

c) A second particle Q also moves along the x-axis so that its velocity is given by :

[tex]V_Q(t)=35\sqrt{t}\cos(0.06t^2)\text{ meters per hour for }0\leq t\leq 10.[/tex]

Hence, the time interval during which the velocity of a particle is atleast 60 meters per hour is [9.404, 10].

Now, the time periods during which a particle's velocity is at least 40 metres per hour are [1.321,4.006] and [9.218, 10]. The distance travelled by the particle Q when its velocity is at least 40 metres per hour is then calculated. :

[tex]\int_{1.321}^{4.006}v_Q(t)dt+\int_{9.218}^{10}v_Q(t)dt\\\\=\int_{1.321}^{4.006}35\sqrt{t}\cos(0.06t^2)dt+\int_{9.218}^{10}35\sqrt{t}\cos(0.06t^2)dt[/tex]

d) At time t = 0, particle Q is is at position x = -90.

We know that P is at xp =  115.7 meters.

Now, The position of Q at t = 10 hr is xq:

[tex]x_q=-90+\int_{0}^{10}v_Q(t)dt=-90+\int_{0}^{10}35\sqrt{t}\cos(0.06t^2)dt[/tex]

And the distance between Q and P is given by :

[tex]|x_p-x_q|=|115.7-(-90+\int_{0}^{10}35\sqrt{t}\cos(0.06t^2)dt)|[/tex]

               [tex]\\=|205.7-\int_{0}^{10}35\sqrt{t}\cos(0.06t^2)dt|[/tex]

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Answer:

Step-by-step explanation: 4(2x-1)+3(2x+5)

Expand the expression to eliminate the brackets

8x-4+6x+15.....Expansion

Now simplify by grouping like terms

8x+6x-4+15

8x+6x=14x

-4+15=11

Therefore simplification=14x+11

Answer:

132.25 π

Step-by-step explanation:

The formula for circumference is 2πr. 2πr = 23π, so r = 11.5

Formula for area is πr^2

11.5^2 * π = 132.25 π

Hope this helps!

In the following question, among the conditions given, {q1, q2} is a basis for the vector space of polynomials p(t) of degree at most two that satisfy the constraint p(2) = 0. In this particular case, we must enter our basis as [[1,0,-4],[0,1,-2]], since q1(t) = t^2 - 4 and q2(t) = t - 2.

To find a basis for the vector space of polynomials p(t) of degree at most two which satisfy the constraint p(2)=0, we can take the following steps:
1. Rewrite the polynomials as linear combinations of the form a + bt + ct^2
2. Use the constraint p(2) = 0 to eliminate one of the coefficients a, b, or c
3. Normalize the polynomials so that they are unit vectors
For example, if your basis is 1 + 2t + 3t^2, 4 + 5t + 6t2 then you can enter it as [[1,2,3],[4,5,6]].

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The hypοtenuse's length is c = 17.

Add the square rοοts οf the οther sides tο find the hypοtenuse. Tο find the shοrter side, subtract the squares οf the οther sides, then take the square rοοt.

Using the Pythagοrean theοrem, we can calculate the length οf the right triangle's missing side:

[tex]a^2 + b^2 = c^2[/tex]

where a, b, and c are the lengths οf the triangle's legs, and c is the length οf the hypοtenuse.

The lengths οf the twο legs are given in this case: a = 8 and b = 15. Sο we can plug the fοllοwing values intο the equatiοn:

[tex]8^2 + 15^2 = c^2[/tex]

[tex]64 + 225 = c^2[/tex]

[tex]289 = c^2[/tex]

When we take the square rοοt οf bοth sides, we get:

[tex]c = \sqrt{(289)} = 17[/tex]

As a result, the hypοtenuse length is c = 17.

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Answer:

the test does not provide convincing evidence that the proportion is greater than 90%

(a)The amount of work needed to stretch the spring from 30 cm to 32 cm is 0.6 J. (b) The distance the spring will be stretched by a 20 N force is 0.03 m.

The formula for the force needed to keep a spring stretched beyond its natural length is F = kx where F is the force, k is the spring constant, and x is the distance from the spring's natural length. The spring constant k is given by the formula: k = (Wd)/x² where W is the work done, d is the distance the spring is stretched from its natural length, and x is the distance from the spring's natural length.

Substituting the values for W, d, and x gives: k = (6 J)/(0.10 m)²

k = 600 N/m

Using the formula F = kx and substituting the values for F and k gives: 20 N = (600 N/m)x

Solving for x gives: x = (20 N)/(600 N/m)

x = 0.0333 m.

Hence, the correct answer is 0.03 m.

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Answer:

100 minutes

Step-by-step explanation:

We know

It takes 3 hrs to run a distance of 180 km.

180 / 3 = 60 km / h

60 minutes = 60 km

40 minutes = 40 km

What time it will take to run 100 km?

60 + 40 = 100 minutes

So, it takes 100 minutes to run 100 km.