THIS IS EASY, BUT I NEED WORK SHOWN, PLEASE HELP!

Mathematics

2 answers:

torisob [31]3 years ago

8 0

Answer:

Step-by-step explanation:

Vlada [557]3 years ago

5 0

Here's how you can simplify the equation:

First off we need to simplify √12x^8, and write it as (√3x^2)(√4x^6),

We could then find the answer:

$\frac{ \sqrt{ {12x}^{8} } }{ \sqrt{ {3x}^{2} } } \\ = \frac{ \sqrt{ {3x}^{2} } \times \sqrt{{4x}^{6} } }{ \sqrt{ {3x}^{2} } } \\ = \sqrt{ {4x}^{6} } \\ = {2x}^{3}$
Thus the answer is 2x^3.

I notice that one of your answers (C,D) are repeated,perhaps you typed something wrong? The answer is either C or D in that case,do tell me if you cannot find 2x^3 at all tho.

Hope it helps!

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a lemon pie was cut into 6 equal pieces. being on a diet, Matilda ate only half a piece. what fraction of the whole pie did she

WINSTONCH [101]

Matilda only ate 1/12 of the pie as 1/2 of 1/6 = 1/12

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3 years ago

Use proper probability notation. Identify function and label inputs when using technology.

zhannawk [14.2K]

Answer:

0.1349

Step-by-step explanation:

Given that:

Sample size, n = 500

20% of 500 ; 0.2 * 500 = 100

p = 0.18 ; n = 500 ; 1 - p = 0.82

P(x ≥ 100) ;

Using the binomial probability relation :

P(x =x) = nCx * p(x)^x * (1 - p)^(n - x

P(x ≥ 100) = 500C100 * 0.18^100 * 0.82^400

P(x ≥ 100) = 0.1349

5 0

3 years ago

If <img src="https://tex.z-dn.net/?f=%20300cm%5E%7B2%7D%20" id="TexFormula1" title=" 300cm^{2} " alt=" 300cm^{2} " align="absmid

Artist 52 [7]

Check the picture below. Recall, is an open-top box, so, the top is not part of the surface area, of the 300 cm². Also, recall, the base is a square, thus, length = width = x.

$\bf \textit{volume of a rectangular prism}\\\\
V=lwh\quad 
\begin{cases}
l = length\\
w=width\\
h=height\\
-----\\
w=l=x
\end{cases}\implies V=xxh\implies \boxed{V=x^2h}\\\\
-------------------------------\\\\
\textit{surface area}\\\\
S=4xh+x^2\implies 300=4xh+x^2\implies \cfrac{300-x^2}{4x}=h
\\\\\\
\boxed{\cfrac{75}{x}-\cfrac{x}{4}=h}\\\\
-------------------------------\\\\
V=x^2\left( \cfrac{75}{x}-\cfrac{x}{4} \right)\implies V(x)=75x-\cfrac{1}{4}x^3$

so.. that'd be the V(x) for such box, now, where is the maximum point at?

$\bf V(x)=75x-\cfrac{1}{4}x^3\implies \cfrac{dV}{dx}=75-\cfrac{3}{4}x^2\implies 0=75-\cfrac{3}{4}x^2
\\\\\\
\cfrac{3}{4}x^2=75\implies 3x^2=300\implies x^2=\cfrac{300}{3}\implies x^2=100
\\\\\\
x=\pm10\impliedby \textit{is a length unit, so we can dismiss -10}\qquad \boxed{x=10}$

now, let's check if it's a maximum point at 10, by doing a first-derivative test on it. Check the second picture below.

so, the volume will then be at $\bf V(10)=75(10)-\cfrac{1}{4}(10)^3\implies V(10)=500 \ cm^3$