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3 years ago

Plzzzzzzzzzzzzzzzzzzzzzzzzzz help quick

Mathematics

1 answer:

SCORPION-xisa [38]3 years ago

5 0

It is C, because the x values between the first dot and the second is 4. So you multiply the slope 4 times on the y value. The y value changes from -2 to -5, so -2 + 4*-3/4 should be equal to -5

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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

Jason ha 2,057 stones in his collection.

Rashid [163]

Answer:

2560 stones

Step-by-step explanation:

Jason has 2057 stones

Paul's stones = Jason + 503

= 2057 + 503 = 2560

5 0

3 years ago

Can someone help me? I do not understand. She did NOT teach us this??

Flauer [41]

I think u just subtract 35° from 130° and that should be the angle of x

7 0

3 years ago

In a jar of ten beads, seven are red and three are blue. a bead is drawn from the jar five times with replacement. what is the p

MatroZZZ [7]

<span>In a jar of ten beads; since 3 are blue; probability of picking a blue ball, B, = p(b) = 3/10. And P (of not picking a blue ball) ; p(b') = 7/10. Since it occurs with replacement, probabilities doesn't change Probaility of picking k blue balls from on n attempts is given by P_n(k) P_n(k) = (n, k) p^(k) q^(n -k) where p and q are b and b' respectively. P_5(2) = (5 , 2) (0.3)^(2) (0.7)^(5 - 2) P_5(2) = 5C2 (0.3)^(2) (0.7)^(3) = 0.3087</span>

6 0

4 years ago

Which one would it be and why.

SVETLANKA909090 [29]

Answer:B because its the rule in algebra that you must gave 1 as a zero

Step-by-step explanation:

5 0

3 years ago