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Ivanshal [37]
2 years ago
15

Point c is the center of dilation. line segment b a is dilated to create line segment b prime a prime. the length of c a is 4. t

he length of a a prime is 16. what is the scale factor of the dilation of line segment ba? one-fifth one-fourth 4 5
Mathematics
1 answer:
pychu [463]2 years ago
8 0

The scale factor of the dilation of line segment ba for which the center of dilation is C, is 5.

<h3>What is the dilation of the figure?</h3>

Dilation of a figure, means the transformation of the figure. The factor by which the given figure dilated, called the scale factor of dilation.

Point c is the center of dilation. Line segment BA is dilated to create line segment b prime, a prime.  The image of the given problem is attached below. The length of CA is 4.

CA=4

The length of a prime is 16.

AA'=16

The new dimension will be,

CA'=CA+AA'\\CA'=4+16\\CA'=20

Now, the scale factor is the ratio of the new dimension to the original dimension. Thus, the scale factor of the dilation is,

r=\dfrac{20}{4}\\r=5

This will be the same for the line segment BA. Thus, the scale factor of the dilation of line segment ba for which the center of dilation is C, is 5.

Learn more about the dilation of the figure here;

brainly.com/question/3457976

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Answer: Hi!

To find the answer for question 24, we would first find the square root of 20 and 5. The square root of 20 is about 4.47 (I used a calculator for convenience) and the square root of 5 is about 2.24. Now we need to add these:

4.47 + 2.24 = 6.71

Now we divide 30 by this:

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Now, we'll evaluate the answer choices.

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7 0
3 years ago
Which is the approximate solution to the system y = 0.5x + 3.5 and y = − 2/3 x + 1/3 shown on the graph? (–2.7, 2.1) (–2.1, 2.7)
AlekseyPX

Answer:

The approximate solution to the system is (-2.7, 2.1).

Step-by-step explanation:

To solve the system of equations \begin{bmatrix}y=0.5x+3.5\\ y=-\frac{2}{3}x+\frac{1}{3}\end{bmatrix} you must:

\mathrm{Rationalize\:equations}\\\\\begin{bmatrix}y=\left(\frac{1}{2}\right)x+\left(\frac{7}{2}\right)\\ y=-\frac{2}{3}x+\frac{1}{3}\end{bmatrix}

\mathrm{Subsititute\:}y=-\frac{2}{3}x+\frac{1}{3}\\\\\begin{bmatrix}-\frac{2}{3}x+\frac{1}{3}=\frac{1}{2}x+\frac{7}{2}\end{bmatrix}

\mathrm{Isolate}\:x\:\mathrm{for}\:-\frac{2}{3}x+\frac{1}{3}=\frac{1}{2}x+\frac{7}{2}\\\\-\frac{2}{3}x=\frac{19}{6}+\frac{1}{2}x\\\\-\frac{7}{6}x=\frac{19}{6}\\\\6\left(-\frac{7}{6}x\right)=\frac{19\cdot \:6}{6}\\\\-7x=19\\\\x=-\frac{19}{7}\approx-2.7

\mathrm{For\:}y=-\frac{2}{3}x+\frac{1}{3}\\\\\mathrm{Subsititute\:}x=-\frac{19}{7}\\\\y=-\frac{2}{3}\left(-\frac{19}{7}\right)+\frac{1}{3}\\\\y=\frac{15}{7}\approx 2.1

The approximate solutions to the system of equations are:

x=-2.7 ,\:y=2.1

5 0
3 years ago
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Please help!! 50/25 points!!<br><br> Compute the sum
AfilCa [17]

Answer:

3577

Step-by-step explanation:

From the question given above, the following data were obtained:

7•2ᶦ

i = 0, 1, 2, .., 8

Sum =?

The sum can be obtained as follow:

7•2ᶦ

i = 0

7•2⁰ = 7 × 1 = 7

i = 1

7•2ᶦ = 7•2¹ = 7 × 2 = 14

i = 2

7•2ᶦ = 7•2² = 7 × 4 = 28

i = 3

7•2ᶦ = 7•2³ = 7 × 8 = 56

i = 4

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i = 5

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i = 6

7•2ᶦ = 7•2⁶ = 7 × 64 = 448

i = 7

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i = 8

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Sum = 7 + 14 + 28 + 56 + 112 + 224 + 448 + 896 + 1792

Sum = 3577

4 0
3 years ago
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