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3 years ago

Which choices correctly describe reflections in the diagram? Check all that apply.

Mathematics

2 answers:

Svet_ta [14]3 years ago

5 0

Answer:

the first one

Step-by-step explanation:

Lunna [17]3 years ago

5 0

Answer:

A.) The red triangle A is reflected across the x-axis onto triangle B.

B.) ✔The red triangle A is reflected across the y-axis onto triangle D.

C.) ✔The blue triangle D is reflected across the x-axis onto triangle C.

D.) ✔The blue triangle D is reflected across the line y = x onto triangle B.

E.) The green triangle C is reflected across the line y = –x onto triangle A.

Step-by-step explanation:

So the answer is <u>B,C, and, D.</u>

Hope this helps

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Paloma can run the 100-meter dash in 16 1/5 seconds. Savannah best time is 19.8 seconds. How much faster is Paloma than Savannah

faust18 [17]

16 1/5 seconds = 16.2 seconds

19.8 - 16.2 = 3.6

Paloma is 3.6 seconds faster than Savannah in the 100-meter dash.

6 0

4 years ago

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

What is 21000 take away two thirds

NISA [10]

Answer:

7000

Step-by-step explanation:

21000/0.667

=14000

21000-14000

=7000

3 0

3 years ago

Can someone please answer this question please answer it correctly and please show work please help me I need it

Annette [7]

4 + -6 = -2 the lines go from 0 to 4 so + 4 then back 6 spots to -2 so 4 + -6 = -2

5 0

3 years ago

1 pts

balu736 [363]

9514 1404 393

Answer:

$935.11

Step-by-step explanation:

The amount is given by the formula ...

A = P(1 +r/n)^(nt) . . . P invested at rate r for t years compounded n per year

A = $850(1 +0.024/2)^(2·4) = $935.11

The amount accumulated will be $935.11 after 4 years.

8 0

3 years ago