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3 years ago

92 POINTS PLS HELP ASAP. I'LL MARK BRAINLIEST

Mathematics

2 answers:

neonofarm [45]3 years ago

6 0

Answer:

Step-by-step explanation:

Its easy

masya89 [10]3 years ago

4 0

Answer:

C is the most reasonable answer because to add the system of equations together, you must make sure that one of the terms (whether it be the x or y value) can cancel out.

Step-by-step explanation:

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I dont get this please help me on this thanks

marishachu [46]

Answer:

x=-2

Step-by-step explanation:

-4(x-7)=36

divide both sides by four

x-7=-9

move constant to the right and change its sign

x=-9+7

calculate the sum

x=-2

7 0

3 years ago

Read 2 more answers

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

Cos θ = -3/5; θ is in QIII. What is sin θ? 4/5 -5/3 -4/5 5/3

Tatiana [17]

Hello :
Cos θ = -3/5
sinθ = - 4/5
because : Cos² θ +sin²θ = (-3/5)²+(-4/5)2 =9/25+16/25 = 25/25 = 1
and θ is in QIII.where : sinθ < 0

3 0

3 years ago

HELP ASAP WILL MARK BRAINLIEST

Jet001 [13]

They lost 2 yards is the answer. But its difficult to tell because it could also be they gained -2 yards.

5 0

3 years ago

Read 2 more answers

169%of 705.8 how would u work that out

Verdich [7]

Answer:

'Percentage' is obtained by multiplying 705.8 by 169%.

169% × 705.8 =

(169 ÷ 100) × 705.8 =

(169 × 705.8) ÷ 100 =

119,280.2 ÷ 100 =

1,192.802 ≈

1,192.8;

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers