FOR 15 POINTS!! Select one of the Theorems from section 2.2 and do the following:
1. Explain why you chose to explore that theorem.
2. Write down the formal definition of the theorem.
3. Explain the theorem in your own words.
4. Find or create an example with new numbers and explain how/why it works.




Here are the 4 Theorems you can choose from:
1. Angle Sum Theorem
2. Third Angle Theorem
3 Exterior Angle Theorem
4. Corollary of Exterior Angle Theorem

Answer: There is a 5.32544378% Chance of a face card being pulled twice

Step-by-step explanation: If there is 52 cards in a deck, and 12 of them are face cards, there is roughly a 23% (23.0769%) chance of pulling one in the first draw. Multiply .230769 x .230769 and you get .0532544378 which equals 5.32544378%

Answer:

Find f(g(x)) f(x)=7x-8 , g(x)=3x-2. f(x)=7x−8 f ( x ) = 7 x - 8 , g(x)=3x−2 g ( x ) = 3 x - 2. Step 1. Set up the composite result function. f(g(x)) f ( g ...

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A catapult, seatbelt, parachute, safety net, or unbalanced force can be used to stop moving objects.

When a seatbelt is fastened during sudden braking or a crash, it exerts pressure on the passenger's body and prevents them from moving forward.

Similar to this, when a parachute is opened, a significant amount of air resistance force is generated, slowing the wearer's descent.

A safety net exerts force on an object that is falling on it as well, absorbing the kinetic energy and stopping it.

A catapult can apply a significant unbalanced force that can halt an object in motion by quickly transferring energy from its springs or elastic materials to the target.

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Answer: Below.

Step-by-step explanation:

To find each we must use the left side to put the first letter. The top for the right letter.

First row - AA

Second row - CB

Third row - CD

Fourth row - BD

A)  The mean distance walked by the group of office workers on Monday was approximately 6.89 km.

b) The interquartile range was approximately 2.2 km.

c)   The value of r would be: r ≈ 4.1 km

d)  (i) Total distance walked = 1.7 km * 20 = 34 km

(ii),  the combined mean distance walked by all thirty workers was approximately 3.28 km.

(iii) the standard deviation is not given directly, we can use the fact that the standard deviation is equal to the square

(a) The mean distance walked by the group of ten office workers on Monday can be found by adding up all of their distances and dividing by the number of workers:

Mean = (5 + 2 + 3 + 6 + 5 + 8 + 5 + 7 + 6 + 4) / 10

= 51 / 10

= 5.1 km

(b) (i) The value of p is the median of the data set. Since there are 10 data points, the median is the average of the 5th and 6th smallest values:

p = (5 + 5) / 2

= 5 km

(c) The value of r represents the maximum distance that is not considered an outlier. From the box-and-whisker plot, we can see that the maximum value is 8 km, which is not an outlier. Therefore,

r = 8 km

(d) (i) The total distance walked by the group of 20 freelancers can be found by multiplying their mean distance by the number of freelancers:

Total distance = Mean distance * Number of freelancers

= 1.7 km * 20

= 34 km

(d)

(i) The total distance walked by the group of 20 freelancers can be found by multiplying their mean distance by the number of freelancers:

Total distance = Mean distance * Number of freelancers

= 1.7 km * 20

= 34 km

(ii) The combined mean distance that all 30 workers walked on Monday can be found by adding up the total distance walked by all workers and dividing by the total number of workers:

Combined mean distance = (Total distance by office workers + Total distance by freelancers) / Total number of workers

= (51 km + 34 km) / 30

= 85 / 30

= 2.83 km

(iii) On Tuesday, all 20 freelancers walked double the distance they walked on Monday. The variance in the distance walked by the freelancers on Tuesday can be found using the formula for the variance of a constant times a random variable:

Var(aX) = a^2 Var(X)

Since the freelancers walked double the distance, the variance of the distance they walked on Tuesday is:

Var(Tuesday distance) = 4 * Var(Monday distance) = 4 * (1 km)^2 = 4 km^2

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Answer: The probability of drawing a vowel from a deck of 12 cards is 5/12, since there are 5 vowels (A, E, I, O, U) and 12 cards in total.

Since Brandi returns the first card drawn to the deck before drawing the second card, the outcome of the first draw does not affect the outcome of the second draw. Therefore, we can treat the two draws as independent events.

The probability of drawing a vowel on the first draw is 5/12, and the probability of drawing a vowel on the second draw is also 5/12. Since we want to find the probability of both events happening (drawing a vowel both times), we can multiply the probabilities:

P(vowel on both draws) = P(vowel on first draw) * P(vowel on second draw)

= (5/12) * (5/12)

= 25/144

Therefore, the theoretical probability that Brandi draws a card with a vowel both times is 25/144, which is approximately 0.1736 or 17.36%.

So the closest answer choice is B) 2.08%.

Step-by-step explanation:

Answer:

80%

Step-by-step explanation:

each side of a square is 30 cm.

decreased by 20%.

find the percentage decrease in the area of the square.

20%x4side= 80%

The number of 5 fluid ounce portions needed in 20 gallons is a total of 512 portions

To solve this problem, we first need to convert the volume of the soup recipe from gallons to fluid ounces, and then divide by the size of each portion to find the total number of portions.

Given that

1 gallon = 128 fluid ounces

Therefore, the recipe yields:

20 gallons x 128 fluid ounces per gallon = 2560 fluid ounces

Now, we can find the number of 5-fluid ounce portions by dividing the total volume by the portion size:

2560 fluid ounces / 5 fluid ounces per portion = 512 portions

Therefore, the soup recipe yields 512 portions of 5 fluid ounces each.

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The derivative of the function g(t) =  -2tsint - sint + 2tcost.

In mathematics, the derivative of a function of a real variable measures the sensitivity of the function's value (the output value) to changes in its independent variable (the input value). Derivatives are a fundamental tool of calculus. The process of finding derivatives is called differentiation. The reverse process is called retro differentiation.

The fundamental theorem of calculus associates inverse differentiation with integration. Differentiation and integration form the two basic operations in univariate calculus.

Given that:

g(t) = tan(ln(t)) g'(t)

Now, Differentiating for the first time:

  (t²+1) d/dt (sin t) + Sin t d/dt (t²+1)

= (t²+1) cos t + sint (2t)

Now,

h'(t) = t²cos t + cost + 2tsint

Differentiating again:

h''(t) = d/dt (t²cos t) + d/dt (cost) + d/dt (2tsint)

      = -2tsint - sint + 2tcost

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If λ is an eigenvalue of A, then λ is an eigenvalue of [tex]A^T[/tex]. Show with an example that the eigenvectors of A and [tex]A^T[/tex] are not the same.

The equation Av = λv, where v is a non-zero vector, is satisfied by an eigenvector v and an eigenvalue given a square matrix A. In other words, the eigenvector v is multiplied by the matrix A to produce a scalar multiple of v. Due to their role in illuminating the behaviour of linear transformations and differential equation systems, eigenvectors play a crucial role in many branches of mathematics and science. When the eigenvector v is multiplied by A, the eigenvalue indicates how much it is scaled.

The eigenvalue and eigenvector states that, let v be a non-zero eigenvector of A corresponding to the eigenvalue λ.

Then, we have:

Av = λv

Taking transpose on both sides we have:

[tex]v^T A^T = \lambda v^T[/tex]

The above equations thus relates transpose of vector and transpose of A to λ.

Now, consider a matrix:

[tex]\left[\begin{array}{cc}1&2\\3&4\\\end{array}\right][/tex]

Now, the eigen values of this matrix are λ1 = -0.37 and λ2 = 5.37.

The eigenvectors are:

[tex]v1 = [-0.8246, 0.5658]^T\\v2 = [-0.4159, -0.9094]^T[/tex]

Now, for transpose of A:

[tex]A^T=\left[\begin{array}{cc}1&3\\2&4\\\end{array}\right][/tex]

The eigen vectors are:

[tex]u1 = [-0.7071, -0.7071]^T\\u2 = [0.8944, -0.4472]^T[/tex]

Hence, we see that, if λ is an eigenvalue of A, then λ is an eigenvalue of [tex]A^T[/tex]. Show with an example that the eigenvectors of A and [tex]A^T[/tex] are not the same.

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There are 4 terms in the series.

what is an arithmetic sequence?

An arithmetic sequence is a sequence of numbers where each term is equal to the previous term plus a fixed constant difference, called the common difference.

The given series is an arithmetic sequence with a common difference of 16. We can find the first term by plugging in n=1 into the sequence:

a1 = -4 + 16(-1)^1 = -4 + 16 = 12

Using the formula for the sum of an arithmetic sequence, we have:

Sn = n/2(2a1 + (n-1)d)

where n is the number of terms, a1 is the first term, and d is the common difference. Plugging in the given values, we have:

3276 = n/2(2(12) + (n-1)(16))

Simplifying this equation, we get:

3276 = n/2(28 + 16n - 16)

3276 = n/2(12 + 16n)

Multiplying both sides by 2 and rearranging, we get:

6544 = n(6 + 8n)

Dividing both sides by 2, we get:

3272 = n(3 + 4n)

We can see that n must be an even number, since the left side is even. We can also see that n cannot be too large, since the right side increases much faster than the left side. Trying some even values of n, we find that:

n=8 -> 3272 = 8(3 + 4(8)) is false

n=6 -> 3272 = 6(3 + 4(6)) is false

n=4 -> 3272 = 4(3 + 4(4)) is true

Therefore, there are 4 terms in the series.

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[tex](\stackrel{x_1}{-6}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{-8}~,~\stackrel{y_2}{-4}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{-4}-\stackrel{y1}{(-3)}}}{\underset{\textit{\large run}} {\underset{x_2}{-8}-\underset{x_1}{(-6)}}} \implies \cfrac{-4 +3}{-8 +6} \implies \cfrac{ -1 }{ -2 } \implies \cfrac{ 1 }{ 2 }[/tex]

[tex]\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{ \cfrac{ 1 }{ 2 }}(x-\stackrel{x_1}{(-6)}) \implies y +3 = \cfrac{ 1 }{ 2 } ( x +6) \\\\\\ y+3=\cfrac{ 1 }{ 2 }x+3\implies {\Large \begin{array}{llll} y=\cfrac{ 1 }{ 2 }x \end{array}}[/tex]

Answer: 90$

Step-by-step explanation:  50% of 60 is 30 so 60+30=90

We can see here that in order to construct a triangle PQR in which angle Q = 30°, angle R=60° and PQ + QR + RP = 10cm​, here is a guide:

A triangle is a geometric shape that is defined as a three-sided polygon, where each side is a line segment connecting two of the vertices, or corners, of the triangle. The interior angles of a triangle always add up to 180 degrees.

Triangles can be classified into different types based on their side lengths and angles, such as equilateral triangles with three equal sides and three equal angles, isosceles triangles with two equal sides and two equal angles, and scalene triangles with no equal sides or angles.

Triangles are used in many areas of mathematics and science, including geometry, trigonometry, and physics.

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The answer of  the given question based on finding each measure of a circle the answer is , (a)  m(MNP) = 12.5° degrees , (b) m(KL) = 102.5° degrees , (c)  m(KJ) = 52.5° degrees , (d)  m(JN) = 102.5° degrees , (e)  m(JLM) =  12.5° degrees.

In geometry,  arc is a portion of  curved line that can be thought of as  segment of  circle. It is defined by two endpoints on  circle and the arc itself is the part of  circle between those two points. An arc can be measured in degrees, and its measure is equal to  central angle subtended by  arc. The length of  arc can also be calculated using the formula L = rθ, where L is  length of  arc, r is radius of  circle, and θ is  angle (in radians) subtended by  arc at center of circle.

Using the properties of angles and arcs in circles:

a. Angle MNP is inscribed in arc MP, so m(MNP) = 1/2m(MP) = 1/2(25) = 12.5° degrees.

Since angles NPM and MPK are vertical angles, we have m(MPK) = m(NPM) = m(MNP) = 12.5° degrees. Then, m(MN) = m(MPK) + m(KPM) = 12.5 + 40 = 52.5° degrees.

b. Angle LKP is inscribed in arc LP, so m(LKP) = 1/2m(LP) = 1/2(25) = 12.5° degrees.

Since angles PKL and LKN are vertical angles, we have m(PKL) = m(LKN) = m(LKP) = 12.5° degrees. Then, m(KL) = m(PKL) + m(PKC) = 12.5 + 90 = 102.5° degrees.

c. Angle PKJ is inscribed in arc PJ, so m(PKJ) = 1/2m(PJ) = 1/2(25) = 12.5° degrees.

Since angles LPK and LPJ are vertical angles, we have m(LPK) = m(LPJ) = m(PKJ) = 12.5° degrees. Then, m(KJ) = m(LPK) + m(LPJ) = 12.5 + 40 = 52.5° degrees.

d. Angle JNM is inscribed in arc JM, so m(JNM) = 1/2m(JM) = 1/2(25) = 12.5° degrees.

Since angles KJM and KJN are vertical angles, we have m(KJM) = m(KJN) = m(JNM) = 12.5° degrees. Then, m(JN) = m(KJM) + m(MJN) = 12.5 + 90 = 102.5° degrees.

e. Angle JLM is an inscribed angle that intercepts arc JM, so m(JLM) = 1/2m(JM) = 1/2(25) = 12.5° degrees.

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Answer: 14.5

Step-by-step explanation:

When there is a decimal point, you start counting from the left any number that is not zero. If the zero is at the end, then you count it.

For example, if the answer is 0.000145 then the number of significant figures is still three because you start counting from the first nonzero number from the left.

If the answer is 14.50, then the number of significant figures is four because you start counting from the first nonzero number from the left.

14.53 is the answer to the equation but because you want to correct it to 3 significant figures, you round down because 3 is less than 5 and 14.5 ends up being the final answer.

With four decimal places added, we have P(2.04 Z 3.04) 0.0189.

To round a decimal value to two decimal places, use the hundredths place, which is the second place to the right of the decimal point.

Subtracting the area to the left of 1.25 from the area to the left of 2.15 will give us the standard normal area between 1.25 and 2.15.

The area to the left of 1.25 is 0.8944, and the area to the left of 2.15 is 0.9842, according to a conventional normal distribution table or calculator.

So, the standard normal area between 1.25 and 2.15 is:

P(1.25 < Z < 2.15) = 0.9842 - 0.8944 = 0.0898

Rounding to four decimal places, we get:

P(1.25 < Z < 2.15) ≈ 0.0898

We follow the same procedure as before to determine the standard normal region between 2.04 and 3.04:

P(2.04 < Z < 3.04) = P(Z < 3.04) - P(Z < 2.04)\

The area to the left of 2.04 is 0.9798, and the area to the left of 3.04 is 0.9987, according to a conventional normal distribution table or calculator.

So, the standard normal area between 2.04 and 3.04 is:

P(2.04 < Z < 3.04) = 0.9987 - 0.9798 = 0.0189

Rounding to four decimal places, we get:

P(2.04 < Z < 3.04) ≈ 0.0189

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Answer:

1

Step-by-step explanation:

[tex]\frac{4}{12}[/tex]  x 3 = 12/12 = 1

Answer:

C = 5n + 4f + 15b

Answer: a

Step-by-step explanation: because a is my name

cos(2a + B) is approximately equal to -0.2948.

why it is?

To find cos(2a + B), we can use the double angle formula for cosine:

cos(2a + B) = cos(2a)cos(B) - sin(2a)sin(B)

We already know the values of a and B, so we can substitute them into the formula and simplify using the trigonometric identities:

a = sin²-1(4/5) = 53.13° (rounded to two decimal places)

B = tan²-1(12/5) = 67.38° (rounded to two decimal places)

cos(2a) = cos²2(a) - sin²2(a) = (cos(a))²2 - (1 - (cos(a))²2) = 2(cos(a))²2 - 1

cos(2a) = 2(sin²-1(4/5))²2 - 1 = 2(0.8)²2 - 1 = 0.32

sin(2a) = 2sin(a)cos(a)

sin(2a) = 2(sin(sin²-1(4/5)))cos(sin²-1(4/5)) = 2(4/5)(3/5) = 0.96

cos(B) = 1/sqrt(1 + tan²2(B)) = 1/sqrt(1 + (12/5)²2) = 5/13

sin(B) = tan(B)cos(B) = (12/5)(5/13) = 0.48

Substituting these values into the formula for cos(2a + B) gives:

cos(2a + B) = cos(2a)cos(B) - sin(2a)sin(B)

cos(2a + B) = (0.32)(5/13) - (0.96)(0.48)

cos(2a + B) = 0.166 - 0.4608

cos(2a + B) = -0.2948

Therefore, cos(2a + B) is approximately equal to -0.2948.

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Answer:

19

Step-by-step explanation:

We are given the following two functions of s

[tex]p(s) = s^3 + 10s\\f(s) = 6s - 3\\\\\text{To find p(2) substitute 2 for s in p(s)}\\p(2) = (2)^3 + 10(2) = 8 + 20 = 28\\\\[/tex]

[tex]\text{To find f(2) substitute 2 for s in f(s)}\\f(2) = 6(2) - 3= 12 - 3= 9\\[/tex]

[tex]p(2) - f(2) = 28 - 9 = 19[/tex]

there is a 20% chance that the point chosen at random will lie on bg.

To find the probability that a point chosen at random will be on the line segment bg, we need to consider the length of bg in relation to the length of the entire line segment ak.

Let us assume that ak is a straight line segment, and bg is a smaller segment that lies entirely within it. To find the probability, we need to divide the length of bg by the length of ak.

Let the length of bg be x and the length of ak be y. Then the probability that a point chosen at random will be on bg is:

Probability = Length of bg / Length of ak

Probability = x / y

However, we need to be careful here. If we choose a point anywhere on ak, it may not necessarily lie on bg. There are an infinite number of points on ak, but only one segment bg. Therefore, the probability we are looking for is actually the ratio of the lengths of bg to ak.

So, if we know the lengths of bg and ak, we can find the probability by dividing them. For example, if bg is 2 units long and ak is 10 units long, the probability of choosing a point on bg is:

Probability = 2 / 10

Probability = 0.2 or 20%

In this case, there is a 20% chance that the point chosen at random will lie on bg.

In conclusion, the probability of a point chosen at random on ak being on bg is directly proportional to the length of bg in relation to the length of ak. Therefore, we need to find the ratio of the lengths of the two line segments to determine the probability.

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what is the probability  a point is chosen at random on ak and then the point will be on bg. dont forget to reduce the products?

Answer:

4 pens.

Step-by-step explanation:

Now he has 2 pens.

Then his friend gives him two more.

As a sum, that's:

2 + 2 = 4 pens.

Answer:

He has 4 pens now.

Step-by-step explanation:

2+2=4

Answer:

87

Step-by-step explanation:

87

Answer:

Independent Variable: Smoking

Quasi-Independent Variable: Parents to Children (Assuming that the researcher purposely manipulated it, otherwise there is none.)

Dependent Variable: Children's attitudes toward smoking behavior

The distance  from Link to the Octorok  is 10.63 units.

We know that the distance between two points (x₁, y₁) and (x₂, y₂) is given by the formula below:

distance = √( (x₂ - x₁)² + (y₂ - y₁)²)

Here we want to find the distance from Link to the Octorok so Link can attack, so we need to get the distance between the points (-4, -5) and (3, 3).

The distance will be:

distance = √( (3 + 4)² + (3 + 5)²)

distance = √( (7)² + (8)²)

distance = √113

distance = 10.63

The distance is 10.63 units.

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Answer:5

Step-by-step explanation:5555555

Answer:

Step-by-step explanation:

D