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3 years ago

Please help ASAP!!!!!! The first figure is dilated to form the second figure

Mathematics

1 answer:

kow [346]3 years ago

5 0

Answer:

the answer is 0.9 is the scale factor

Step-by-step explanation:

id you subtract 1.5 and 0.6 you get 0.9

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which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago

A bag contains tiles number 1-10. What is the probability of first selecting a PRIME number, replacing it, then selecting a numb

andrezito [222]

Answer:

0.12 = 12% probability of first selecting a PRIME number, replacing it, then selecting a number less than or equal to 3

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

Probability of independent events:

Suppose we have two events, A and B, that are independent. The probability of both happening is given by:

$P(A \cap B) = P(A) \times P(B)$

In this question:

Event A: Selecting a prime number.

Event B: Selecting a number less than or equal to 3

Probability of selecting a prime number:

Between 1 and 10, we have 4 prime numbers(2,3, 5 and 7), out of 10. So

$P(A) = \frac{4}{10}$

Probability of selecting a number less than or equal to 3:

Three numbers(1,2,3) out of 10. So

$P(B) = \frac{3}{10}$

Probability of both:

$P(A \cap B) = P(A) \times P(B) = \frac{4}{10} \times \frac{3}{10} = \frac{12}{100} = 0.12$

0.12 = 12% probability of first selecting a PRIME number, replacing it, then selecting a number less than or equal to 3

5 0

2 years ago

What is the starting point of this function

Finger [1]

the starting point of the function is 0

6 0

3 years ago

sleet_krkn [62]

Answer:

Divide x by 3 O using long polynomial division.

Step-by-step explanation:

8 0

2 years ago

Please help click on the image below

Arisa [49]

I think it C
have a great day!!!

3 0

2 years ago

Read 2 more answers