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4 years ago

15

Simplify. Rewrite the expression in the form k⋅y^n. (2y^4)^3

1 answer:

N76 [4]3 years ago

5 0

Answer:

The answer is

<h2> ${8.y}^{12}$ </h2>

Step-by-step explanation:

(2y⁴)³

Using the rules of indices

That's

<h3> $( {ax})^{b} = {a}^{b} . {x}^{b}$ </h3>

So for the above expression we have

<h3> $( {2y^{4} })^{3} = {2}^{3} . {y}^{4 \times 3}$ </h3>

Simplify

We have the final answer as

<h2>8.y¹²</h2>

Hope this helps you

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A solid oblique pyramid has a regular hexagonal base with an area of 54 cm2 and an edge length of 6 cm. Angle BAC measures 60°.W

sergij07 [2.7K]

Answer:

324 cm3

Correct on edge

4 0

3 years ago

Read 2 more answers

Rationalize the denominator of sqrt -9 / (4-7i) - (6-6i) ...?

Colt1911 [192]

√(- 9 ) / (( 4 - 7 i ) - ( 6 - 6 i )) = 3 i / ( 4 - 7 i - 6 + 6 i ) =
= 3 i / ( - 2 - i ) = - 3 i / ( 2 + i ) =
$= \frac{-3i}{2 + i}* \frac{2 - i}{2 - i}= \\ = \frac{-6i+3i ^{2} }{2- i^{2} }= \\ = \frac{-3-6i}{2+1}= \frac{-3-6i}{3}=$
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5 0

4 years ago

Peter Schmidt ordered sandwiches and salads for a luncheon meeting. The sandwiches cost three times as much as a salads. The tot

evablogger [386]

EXPLANATION

Let's see the facts:

$Sandwiches= 3*$Salads

Bill = $24.92

We can represent this in an equation:

Total Bill = Sandwiches + Salads

Let's call to Sandwiches "x" to Salads "y" and to Bill "B"

Then, the equation will be:

B = x+y = (3*y) + y

Applying distributive property:

24.92 = 3y + y

Adding similar terms:

24.92 = 4y

Dividing both sides by 4:

6.23 = y

Switching:

y=6.23

As we know, y variable represent the salads cost, so:

Answer is $6.23.

3 0

1 year ago

How do you simplify <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B2%7D%20%2B%5Csqrt%7B6%7D%20%7D%7B%5Csqrt%7B8%7D%20%2B%

blondinia [14]

The trick is to exploit the difference of squares formula,

$a^2-b^2=(a-b)(a+b)$

Set a = √8 and b = √6, so that a + b is the expression in the denominator. Multiply by its conjugate a - b:

$(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)=(\sqrt8)^2-(\sqrt6)^2=8-6=2$

Whatever you do to the denominator, you have to do to the numerator too. So

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}{(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}2$

Expand the numerator:

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{2\cdot8}+\sqrt{6\cdot8}-\sqrt{2\cdot6}-(\sqrt6)^2$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{16}+\sqrt{48}-\sqrt{12}-6$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=4+\sqrt{48}-\sqrt{12}-6$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(\sqrt4-1)$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(2-1)$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6}=-2-\sqrt{12}$

So we have

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+\sqrt{12}}2$

But √12 = √(3•4) = 2√3, so

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+2\sqrt3}2=\boxed{-1-\sqrt3}$

7 0

3 years ago

R= 10,15,20 S=20,25 R u S =

dolphi86 [110]

The answer to the question

6 0

3 years ago