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3 years ago

Line segment AB is dilated to create line segment A'B' using point Q as the center of dilation.

Mathematics

2 answers:

barxatty [35]3 years ago

8 0

Answer:

Step-by-step explanation:

Rashid [163]3 years ago

7 0

Answer:

Step-by-step explanation:

To find the scale of dilation, you have to compare the ratio of the after transform with before transform according to the center of dilation. In this case, the center of dilation is Q so you have to compare QA' : QA. We know that QA= 1.25 and AA' is 1.25. The distance of QA' will be:

QA'= QA + AA'= 1.25 + 1.25 = 2.5

The scale factor will be:

scale factor= QA' : QA.= 2.5 : 1.25

scale factor= 2

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Enter as a integer or decimal

valentina_108 [34]

Answer:

x = - 6

Step-by-step explanation:

Given

4(x + 6) - 2x = 12 ← distribute and simplify left side

4x + 24 - 2x = 12

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3 0

3 years ago

The graph of y=cos⁡x is transformed to y=a cos ⁡(x−c)+d by a vertical compression by a factor of 1/3 and a translation 2 units d

rodikova [14]

ANSWER

$y = \frac{1}{3} \cos(x) - 2$

EXPLANATION

If the graph of y=cos⁡x is transformed to y=a cos ⁡(x−c)+d by a vertical compression by a factor of 1/3 and a translation 2 units down,

then

a=1/3

and d=-2.

The 'c' is a phase shift since it is not given, it means it is zero.

Therefore the new equation is:

y=1/3cos⁡(x-0)−2

This simplifies to:

y=1/3cos⁡x−2

The correct option is C.

5 0

3 years ago

Read 2 more answers

Explanation also plz

ozzi

Answer:

????

Step-by-step explanation:

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7 0

3 years ago

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mash [69]

Answer:

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Step-by-step explanation:

3 0

3 years ago

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Prove that :( 1 + 1/<img src="https://tex.z-dn.net/?f=tan%5E%7B2%7DA" id="TexFormula1" title="tan^{2}A" alt="tan^{2}A" align="ab

Aliun [14]

Answer:

See explanation

Step-by-step explanation:

Simplify left and right parts separately.

$\left(1+\dfrac{1}{\tan^2A}\right)\left(1+\dfrac{1}{\cot ^2A}\right)\\ \\=\left(1+\dfrac{1}{\frac{\sin^2A}{\cos^2A}}\right)\left(1+\dfrac{1}{\frac{\cos^2A}{\sin^2A}}\right)\\ \\=\left(1+\dfrac{\cos^2A}{\sin^2A}\right)\left(1+\dfrac{\sin^2A}{\cos^2A}\right)\\ \\=\dfrac{\sin^2A+\cos^2A}{\sin^2A}\cdot \dfrac{\cos^2A+\sin^A}{\cos^2A}\\ \\=\dfrac{1}{\sin^2A}\cdot \dfrac{1}{\cos^2A}\\ \\=\dfrac{1}{\sin^2A\cos^2A}$

<u>Right part:</u>

$\dfrac{1}{\sin^2A-\sin^4A}\\ \\=\dfrac{1}{\sin^2A(1-\sin^2A)}\\ \\=\dfrac{1}{\sin^2A\cos^2A}$

Since simplified left and right parts are the same, then the equality is true.

3 0

3 years ago