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2 years ago

Divide square root 9^2 by square root 18y^2

1 answer:

Klio2033 [76]2 years ago

6 0

I got y^2/2

My Solution:

√9^2/18 · y^2

= 1/2 y^2

Multiply Fractions: a · (b/c) = (a · b)/c

1·y^2/2

=y^2/2

Hope I helped,
Faith xx

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9,864,320

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3 years ago

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Valerie had a bottle that was filled with that contained 128oz juice to put into cups, she filled each cup with 6 floz of juice,

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3 years ago

Graph the equation y=-2(x-1)^2-4

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3 years ago

Which expression is equivalent to (16 x Superscript 8 Baseline y Superscript negative 12 Baseline) Superscript one-half?.

loris [4]

To solve the problem we must know the Basic Rules of Exponentiation.

<h2>Basic Rules of Exponentiation</h2>

$x^ax^b = x^{(a+b)}$
$\dfrac{x^a}{x^b} = x^{(a-b)}$
$(a^a)^b =x^{(a\times b)}$
$(xy)^a = x^ay^a$

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

The solution of the expression is $\dfrac{4x^4}{y^6}$ .

<h2>Explanation</h2>

Given to us

$(16x^8y^{12})^{\frac{1}{2}}$

Solution

We know that 16 can be reduced to $2^4$ ,

$=(2^4x^8y^{12})^{\frac{1}{2}}$

Using identity $(xy)^a = x^ay^a$ ,

$=(2^4)^{\frac{1}{2}}(x^8)^{\frac{1}{2}}(y^{12})^{\frac{1}{2}}$

Using identity $(a^a)^b =x^{(a\times b)}$ ,

$=(2^{4\times \frac{1}{2}})\ (x^{8\times\frac{1}{2}})\ (y^{12\times{\frac{1}{2}}})$

Solving further

$=2^2x^4y^{-6}$

Using identity $\dfrac{x^a}{x^b} = x^{(a-b)}$ ,

$=\dfrac{2^2x^4}{y^6}$

$=\dfrac{4x^4}{y^6}$

Hence, the solution of the expression is $\dfrac{4x^4}{y^6}$ .

Learn more about Exponentiation:

brainly.com/question/2193820

8 0

2 years ago

The sum of 4 consecutive integers is 198. What is the fourth number in this sequence.

Georgia [21]

Let x, x+1,x+2, x +3 are the <span>numbers
so
</span>x + x +1 + x +2 +x +3 = 198
4x + 6 = 198
4x =198-6
4x = 192
x = 192/4
x = 48
x + 1 = 48 + 1 = 49
x + 2 = 48 + 2 = 50
x + 3 = 48 + 3 = 51

proof
48 + 49 + 50 + 51 = 198

so the numbers are 48, 49, 50, 51

fourth number in this sequence is 51

7 0

3 years ago