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alexdok [17]
3 years ago
5

Find the scale factor of the dilation. Then tell whether it is enlargement or reduction.​

Mathematics
1 answer:
mariarad [96]3 years ago
4 0

Answer:

  a) 1/2; reduction

  b) 5/4; enlargement

Step-by-step explanation:

In each case, the scale factor is CP'/CP. When it is more than 1, the dilation is an enlargement.

Even before you run the numbers, you can tell if it is an enlargement or not. If the dilated figure is larger, P is closer to C than is P'. If P' is closer to C, then it is a reduction.

a) CP'/CP = (4-2)/4 = 2/4 = 1/2 . . . . a reduction

__

b) CP'/CP = 25/20 = 5/4 . . . . an enlargement

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How is 17 + 26 different from 36 + 53? Explain ( this is regrouping lesson)
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3 years ago
Engineers are planning a new bridge to cross a small river. Surveyors created the diagram to show where the bridge would cross t
Bogdan [553]

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Help with Algebra 2: 1. <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B7%5D%7Bx%5E5%7D%20%7D%7B%5Csqrt%5B4%5D%7Bx%5E2%7D%
Mars2501 [29]

Answer:

See explanation

Step-by-step explanation:

1. Given the expression

\dfrac{\sqrt[7]{x^5} }{\sqrt[4]{x^2} }

Note that

\sqrt[7]{x^5}=x^{\frac{5}{7}} \\ \\\sqrt[4]{x^2}=x^{\frac{2}{4}}=x^{\frac{1}{2}}

When dividing \sqrt[7]{x^5} by \sqrt[4]{x^2}, we have to subtract powers (we cannot subtract 4 from 7, because then we get another expression), so

\dfrac{5}{7}-\dfrac{2}{4}=\dfrac{5}{7}-\dfrac{1}{2}=\dfrac{5\cdot 2-1\cdot 7}{14}=\dfrac{3}{14}

and the result is x^{\frac{3}{14}}=\sqrt[14]{x^3}

2. Given equation 3\sqrt[4]{(x-2)^3} -4=20

Add 4:

3\sqrt[4]{(x-2)^3} -4+4=20+4\\ \\3\sqrt[4]{(x-2)^3}=24

Divide by 3:

\sqrt[4]{(x-2)^3} =8

Rewrite the equation as:

(x-2)^{\frac{3}{4}}=8\\ \\(x-2)^{\frac{3}{4}}=2^3

Hence,

\left((x-2)^{\frac{3}{4}}\right)^{\frac{4}{3}}=(2^3)^{\frac{4}{3}}\\ \\x-2=2^{3\cdot \frac{4}{3}}\\ \\x-2=2^4\\ \\x-2=16\\ \\x-2+2=16+2\\ \\x=18

3 0
3 years ago
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