Why does the base of an exponent have to be greater then 1

1 answer:

7 0

It's not 1 that is exactly the problem. It is also 0. And perhaps - 1 although that is a lot trickier than the other two.

Let's start with one. It give 1^1 = 1 for a power of 1. It gives 1^2 = 1 * 1 for a power of 2. It will give 1^100000 = 1 for a power of 100000.

So the base can be 1 but it is known that any power won't change it's value.

The same can be said of 0. 0^1000 = 0^1 = 0. 0^0 is really a bad dude. You won't learn about that for awhile.

(-1)^11 is different from (-1)^10. The first gives -1 and the second gives 1. So you have to be careful with -1.

There is nothing wrong with (1/4)^3. The power of three and the base of 1/4 are fine. The answer is (1/4)*(1/4) * (1/4) = 1/64 which is perfectly good answer.

It works for negatives too. (-1/2)^3 = - 1/8

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The price of a house is originally listed at $155,000. The owners are having a hard time selling it and decide to reduce the pri

saul85 [17]

Answer:

83%

Step-by-step explanation:

We can put this into a fraction to get the ratio of the new price over the old price

$\frac{128650}{155000}$

This will give us the percentage of how much was decreased.

The answer you get is .83

Bring 2 decimal places to the right to get 83 (converting from decimal to percentage)

You get 83 percent to be your answer

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2 years ago

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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 $\bf V(10)=75(10)-\cfrac{1}{4}(10)^3\implies V(10)=500 \ cm^3$

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KengaRu [80]

Answer:

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What os the domain of the given function

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