Two parallel lines are cut by a transversal as shown below.
Suppose m 2 8 = 132º. Find m Z1 and m 2 3.
2.
4
3
5
6
8
8
7

Answer: 138 inches

Step-by-step explanation:

We can use proportions here, let h = height of the statue

height/height = shadow/shadow

\(\frac{h}{69} = \frac{94}{47}\)

h/69 = 2

*69 (h/69 = 2) *69

h = 138 inches

Answer:

First we need to solve:

3x + 9 < 30

-9         -9

3x < 31

Divide by 3 on both sides:

x < 10.3 around 10

this doesn't really seem right are you sure you typed it in correctly?

Step-by-step explanation:

x < 10 im assuming its 13 i guess since there are no options of 10..

so:

x < 13

B is your answer

The circle is filled in, and it starts at the point of 13 and goes down.

The slopes are best approximated as: The slopes are approximately -1.7, 0, and 1.7.

The formula to find the slope between two coordinates is expressed as:

Slope = (y₂ - y₁)/(x₂ - x₁)

The slope of each line above the x-axis are:

Slope 1 = (5 - 0)/(5 - 8)

Slope 1 = -1.7

Slope 2 = (5 - 5)/(5 - (-2))

Slope 2 = 0

Slope 3 = (5 - 0)/(-2 - (-5))

Slope 3 = 5/3 = 1.7

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This set {σ ∈ S4 : σ(1) = 3} contains the following elements of S4: (13), (23), (34), (24), (14), (12)(34), (13)(24), (14)(23) and  {σ ∈ S4 : σ(2) = 2} this set contains the following elements of S4: (12), (21), (13)(24), (24)(13).

The group S4 is the symmetric group on 4 elements and has 24 elements. We can list all of its subgroups as follows:

Now, let's consider the sets (a) and (b) and see if they are subgroups of S4:

(a) {σ ∈ S4 : σ(1) = 3}

This set contains the following elements of S4: (13), (23), (34), (24), (14), (12)(34), (13)(24), (14)(23). We can check that this set is not a subgroup of S4 because it is not closed under composition. For example, (13)(14) = (134) is not in the set.

(b) {σ ∈ S4 : σ(2) = 2}

This set contains the following elements of S4: (12), (21), (13)(24), (24)(13). We can check that this set is a subgroup of S4. It is closed under composition, inverses, and contains the identity element e. Therefore, it is a subgroup of S4 and is isomorphic to the cyclic group of order 2.

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Answer:

x + x - 45 = 90

2x - 45 = 90

2x = 135

x = 67.5, so x - 45 = 22.5

The other two angles measure 22.5° and 67.5°.

Answer:

633.3

Step-by-step explanation:

Answer:

-5x+6-3x=22 is x = -2

10-9x+6=34 is x = -2

Step-by-step explanation:

Airplane 1 had the fastest rate of descent during its landing.

To compare the rates of descent of the two airplanes, we need to convert both rates to a common unit of measurement. Let's convert both rates to feet per minute (ft/min) to make the comparison easier.

Airplane 1 had a rate of descent of -130 ft/s, which is equivalent to:

-130 ft/s × 60 s/min = -7800 ft/min

Airplane 2 had a rate of descent of -97/8 ft/s, which is equivalent to:

(-97/8) ft/s × 60 s/min = -729.375 ft/min

Comparing these two rates, we can see that Airplane 1 had the faster rate of descent during its landing:

-7800 ft/min (Airplane 1) > -729.375 ft/min (Airplane 2)

Therefore, Airplane 1 had the fastest rate of descent during its landing.

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Answer:

Where is it?

Step-by-step explanation:

Answer:

240°

Step-by-step explanation:

measure of XWZ = 360 - 120 = 240°

The measure of arc wz is 240 degrees.

here, we have,

we know that,

A small arc is one with a measure of fewer than 180 degrees. A major arc is one with a measure greater than 180 degrees. A semicircle is an arc that divides a circle in half and has a measure of 180 degrees.

An arc is a continuous segment of a circle. There are two sections, one of which is larger than the other. The major arc is the larger one, and the minor arc is the smaller one.

Given:

From diagram, m ∠xz = 120 degrees.

We know that whole circle = 360 degrees

So, measure of arc wz = 360 - 120

                                     = 240

Hence, the measure of arc wz is 240 degrees.

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The number of quarters and dimes with Maria are 17 and 8 respectively.

Linear equations in mathematics are ones where the greatest power of the variable is one. A linear equation can only have one possible solution: a straight line. Thus, a simultaneous linear equation is a system of two linear equations in two or three variables that are solved concurrently to get a shared answer. Three methods are frequently used to solve simultaneous linear equations: substitution method, elimination method, and graphic method.

Let q be the number of quarters.

Let d be the number of dimes.

Given that the total amount is 5.05 in quarters and dimes

Therefore, 0.25 q + 0.10d = 5.05;

I once more, q = 2d + 1. — (ii) (ii)

Inferring from I and (ii), d = 4.80/0.60 = 8 and 0.25(2d + 1) + 0.10d = 5.05 and 0.60d = 4.80 respectively.

This is put into (ii), and the result is q = 2d + 1 = 2*8 + 1 = 17.

As a result, q = 17 and d = 8

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The given diagram implies that the triangle $ABC$ is an isosceles triangle with angles of $24^\circ$. This can be verified by computing the length of the hypotenuse and using the Pythagorean Theorem.

Equation is a mathematical expression that consists of variables, symbols, and numbers, and shows the relationship between different quantities. An equation can involve one or more unknowns, and can be represented as an equality or as an inequality. Solving an equation requires understanding the relationship between the different elements in the equation, and manipulating the equation to isolate the unknowns. Common types of equations include linear equations, quadratic equations, and polynomial equations.

From this diagram, we can conclude that $\angle ABC$ is an isosceles triangle. This is because the angles opposite equal sides of a triangle must be equal. Since $\angle BAC=24^\circ$, $\angle ABC$ must be $24^\circ$. Furthermore, the sides $AB$ and $AC$ are equal, so $ABC$ is an isosceles triangle.

This conclusion can be verified by using the Pythagorean Theorem. If $AB=AC$, then it follows that $BC = \sqrt{AB^2 + AC^2} = \sqrt{2AB^2}$. Since $AB=3$, it follows that $BC=\sqrt{2*3^2}=6$. Therefore, the triangle $ABC$ is a right triangle with legs of length 3 and hypotenuse of length 6. Since $\angle BAC = 24^\circ$, it follows that $\angle ABC = 24^\circ$, verifying that the triangle is isosceles.

In conclusion, the given diagram implies that the triangle $ABC$ is an isosceles triangle with angles of $24^\circ$. This can be verified by computing the length of the hypotenuse and using the Pythagorean Theorem.

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Answer:

78

Step-by-step explanation:

Answer:

the answer is C

........................

Answer:

x=-5

Step-by-step explanation:

Answer:

c took me a wiel but then realized it was just butterfly method to solve.

Step-by-step explanation:

The coordinates in the solution to the systems of inequalities is (12, 1)

From the question, we have the following parameters that can be used in our computation:

y > -4x + 6

y > 1/3x - 7

Next, we plot the graph of the system of the inequalities

See attachment for the graph

From the graph, we have solution to the system to be the shaded region

The coordinates in the solution to the systems of inequalities graphically is (12, 1)

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Answer:

7pi/9

Step-by-step explanation:

subtract 360 from 500 as many times as you can

number left should be converted to radians

500 - 360 = 140

140 * (pi/180) =

140Pi/180 =

simplify by dividing by 20 top & bottom

7Pi/9

varsitytutors

Answer:

X= 2/5 y=-2

Step-by-step explanation:

5x + 2y = -2

3x - 5y = 11.2

Using the elimination method

Multiply equation 1 by 3

Multiply equation 2 by 5

15x + 6y = -6

15x - 25y = 56

Subtrat  3 from 4

-31y= 62

Divide both sides by -31

y= -2

To arrive at x, substitute y= -2 in equation 1

5x + 2y = -2

5x + 2(-2)= -2

5x -4 = -2

Collect like terms

5x= -2+4

5x= 2

Divide both sides by 5

x= 2/5

The most appropriate model for this set of data is \(0.5x^2 +20\) (option-c)  

When determining the appropriate model for a set of data, we want to choose the equation that best fits the pattern observed in the data points. In this case, we have the following data: (-1,20), (0, 10), (1, 5), and (2, 2.5).

One possible model is a linear model, such as y = mx + b, where m is the slope and b is the y-intercept. We can calculate the slope and y-intercept for our data set by using linear regression analysis. However, when we plot the data points, it is clear that the data does not follow a linear trend.

Another possible model is a quadratic model, such as y = ax^2 + bx + c. This model assumes that the relationship between the independent variable (x) and the dependent variable (y) is quadratic. Again, when we plot the data points, it appears that the quadratic model is also not an appropriate fit.

The most apparent pattern in the data is a linear decrease in the y-values as the x-values increase. This equation represents a linear relationship, where the y-intercept (when x = 0) is 10, and the slope is -0.5. This equation accurately describes the trend observed in the data and can be used to make predictions for future x-values.(option-c)

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Answer:

4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, Step-by-step explanation:

Therefore, additivity and homogeneity of T are satisfied, T is linear.

T must be linear for both b and c to be true. T is linear, hence additivity is true for every p, q, and P. (R). In order to make our computations as straightforward as feasible, it would be a good idea for us to use straightforward polynomials in P(R). p, q ∈ P(R), where p(x) = \(\frac{\pi }{2}\) and q(x) =  \(\frac{\pi }{2}\) for all x ∈ R and so we have

T(p + q) =(3(p + q)(4) +5(p + q)'(6)+b(p + q)(1)(p + q)(2) , \(\int\limits^2_ {-1} \,\) \(x^{3}\)(p + q)(x) d(x) +               c sin((p + q)(0))

           = (3(p(4)+q(4)) + 5(p'(6)+q'(6))+b(p(1)+q(1))(p(2)+q(2)) , \(\int\limits^2_ {-1} \,\) \(x^{3}\)(p(x)+q(x))d(x)+ c sin(p(0)+q(0)))

           = (3(\(\frac{\pi }{2}\) +\(\frac{\pi }{2}\))+5(0+0)+b(\(\frac{\pi }{2}\)+\(\frac{\pi }{2}\))(\(\frac{\pi }{2}\)+\(\frac{\pi }{2}\)),  \(\int\limits^2_ {-1} \,\) \(x^{3}\)(\(\frac{\pi }{2}\)+\(\frac{\pi }{2}\))d(x)+c sin(\(\frac{\pi }{2}\)+\(\frac{\pi }{2}\)))

           = (3(\(\pi\))+b\(\pi ^{2}\),\(\frac{15\pi }{4}\))

and

Tp + Tq = (3p(4) +5(p)'(6)+bp(1)(p)(2) , \(\int\limits^2_ {-1} \,\) \(x^{3}\)(p)(x) d(x) + c sin((p)(0)) +(3(q)(4) +5(q)'(6)+b(q)(1)(q)(2) , \(\int\limits^2_ {-1} \,\) \(x^{3}\)(q)(x) d(x) +c sin(q(0)))

           = (3(\(\frac{\pi }{2}\))+5(0)+b(\(\frac{\pi }{2}\))(\(\frac{\pi }{2}\)),  \(\int\limits^2_ {-1} \,\) \(x^{3}\)(\(\frac{\pi }{2}\))d(x)+c sin(\(\frac{\pi }{2}\))) +  (3(\(\frac{\pi }{2}\))+5(0)+b(\(\frac{\pi }{2}\))(\(\frac{\pi }{2}\)),  \(\int\limits^2_ {-1} \,\) \(x^{3}\)(\(\frac{\pi }{2}\))d(x)+c sin(\(\frac{\pi }{2}\)))

           =(3(\(\frac{\pi }{2}\))+\(\frac{\pi b}{4}\),\(\frac{\pi }{2}\)  \(\int\limits^2_ {-1} \,\) \(x^{3}\)d(x) + c) +(3(\(\frac{\pi }{2}\))+\(\frac{\pi b}{4}\),\(\frac{\pi }{2}\)  \(\int\limits^2_ {-1} \,\) \(x^{3}\)d(x) + c)

           =(3\(\pi\)+ \(\frac{\pi b}{2}\),\(\frac{15\pi }{4}\)+2c)

Since T is linear, additivity of T holds and implies that we have

              (3(\(\pi\))+b\(\pi ^{2}\),\(\frac{15\pi }{4}\)) = T(p+q)

                                     =Tp+Tq

                                     =(3\(\pi\)+ \(\frac{\pi b}{2}\),\(\frac{15\pi }{4}\)+2c)

from which we can equate the coordinates to obtain the equations 3π + πb/2= 3π +πb/2 and 15π/4 =15π/4+2c, which imply b = 0 and c = 0, respectively.

Backward direction: If b = 0 and c = 0, then T is linear. Suppose b = 0 and c = 0. Then the map T : R ^3 → R^2 becomes

Tp = (3p(4) +5(p)'(6) , \(\int\limits^2_ {-1} \,\) \(x^{3}\)(p)(x) d(x) )

we need to prove that T is linear

• Additivity: For all p, q ∈ P(R), we have

 T(p+q) = (3(p + q)(4) +5(p + q)'(6) , \(\int\limits^2_ {-1} \,\) \(x^{3}\)(p + q)(x) d(x))

             =(3(p(4)+q(4)) + 5(p'(6)+q'(6)) , \(\int\limits^2_ {-1} \,\) \(x^{3}\)(p(x)+q(x))d(x))

             =  (3p(4) +5(p)'(6)+3q(4)+5q'(6) , \(\int\limits^2_ {-1} \,\) \(x^{3}\)(p)(x) d(x) + \(\int\limits^2_ {-1} \,\) \(x^{3}\)q(x) d(x))

              = (3p(4) +5(p)'(6) , \(\int\limits^2_ {-1} \,\) \(x^{3}\)(p)(x) d(x)) +(3(q)(4) +5(q)'(6), \(\int\limits^2_ {-1} \,\) \(x^{3}\)(q)(x) d(x))

             =Tp+Tq

• Homogeneity: For all λ ∈ F and for all (x, y, z) ∈ R^3, we have

                 T(λp) = (3(λp)(4) + 5(λp)'(6), \(\int\limits^2_ {-1} \,\) \(x^{3}\)(λp)(x) d(x))

                           =(3λp(4) + 5λp'(6), \(\int\limits^2_ {-1} \,\) \(x^{3}\)λ(p)(x) d(x))

                          =(λ(3p(4) + 5p'(6)),λ \(\int\limits^2_ {-1} \,\) \(x^{3}\)p)(x) d(x))

                          =λ(3p(4) + 5p'(6), \(\int\limits^2_ {-1} \,\) \(x^{3}\)p)(x) d(x))

                         =λTp

 Therefore, additivity and homogeneity of T are satisfied, T is linear.

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Answer:

with the years labeled on the horizontal axis and the number of hours labeled on the vertical axis. The graph shows that the number of hours spent watching TV declined steadily over the 5-year period, starting at around 1750 hours in Year 1 and falling to around 1300 hours in Year 5. The graph also shows a clear, downward-sloping trend from Year 1 to Year 5.

Step-by-step explanation:

Answer:

Lateral area = 2.pi.r.h = 2 x pi x 2 x 3 = 12 pi

Total area = lateral area + pi. r^2 . 2 = 20 pi

Volume = pi . r^2 . 2 . h = 24 pi

The area of the square hole having a side length of 6 inches is 36 inches².

The area of a square is a product of it's any two sides or a product of diagonals divided by two. If side has a length a then diagonal is a√2.

The perimeter of a square is the sum of the lengths of all the sides.

Given, Darius cuts out a square hole that is 6 inches long each side.

We know the area of a square is (side×side).

Therefore, The area of the square hole having a side length of 6 inches is = (6×6) = 36 inch²

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Answer:

  4 and 5

Step-by-step explanation:

For answering questions like this, it can be useful to remember a few of the powers of small integers:

  2^4 = 16

  2^5 = 32

A logarithm can be considered to be an exponent of the base.

  \(\log_b(x) = a \ \Longleftrightarrow\ b^a=x\)

The ordering of powers of 2 relative to the number of interest (17) is ...

  16 < 17 < 32

  2⁴ < 17 < 2⁵ . . . . . . . . . . . . . . . . . . . expressed as powers of 2

  log₂(2⁴) < log₂(17) < log₂(2⁵) . . . . . log₂ of the above inequality

  4 < log₂(17) < 5 . . . . . . . . . . . . . . . . showing the values of the logs

Log₂(17) lies between 4 and 5.

__

Additional comment

Using the "change of base" formula, you can use a calculator to find the value of log₂(17). It shows you the value is between 4 and 5.

  log₂(17) = log(17)/log(2) . . . . . . using logs to the same base

Answer:

B

Step-by-step explanation:

edge2020

Answer:

B.

Step-by-step explanation: EDGE 2020

The scaled triangle will be larger than the initial size by a factor 2.

The scaled square will be smaller than the initial side by a factor 4.

Dilation refers to a transformation that changes the size of a geometric figure without altering its shape.

Dilation involves scaling an object by a certain factor, that might result in  enlarging or reducing its dimensions uniformly in all directions.

Based on the given diagram, the new length and size of the object is calculated as follows;

For the triangle, (measure the length with ruler)

For the square; (measure the length with ruler)

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