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2 years ago

How would you write 8^5 as a multiplication expression? A. 8 × 5 B. 8 × 8 × 8 × 8 × 8 C. 5 × 5 × 5 × 5 × 5 × 5 × 5 × 5

Mathematics

2 answers:

Helga [31]2 years ago

7 0

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$\diamond\:\large\blue\textsf{\textbf{\underline{\underline{Question:-}}}}\:\diamond$

✶ How would you write $\bold{8^5}$ as a multiplication expression $\textit{?}$

$\diamond\large\blue\textsf{\textbf{\underline{\underline{Answer and How to solve:-}}}}\diamond$

✶ The exponent indicates how many times a certain number was multiplied by itself.

✶ In this case we have $\bold{8^5}$ , so 8 was multiplied by itself five times.

So we conclude that $\bold{8^5}$ written as a multiplication expression looks as follows:-

$\bigcirc\!\!\!\!\checkmark$ 8•8•8•8•8

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Neporo4naja [7]2 years ago

4 0

$\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}$

if we ever face a number written in the form of $\underline{x^{n}}$ where x denotes the base and n denotes the exponent or power , we can expand it in the following way -

$x^{n} = x \times x \times x \times.... x ( upto\: " n " \: times )$

therefore ,

$8 {}^{5} = 8 \times 8 \times 8 \times 8 \times 8$

option ( B )

hope helpful -,-

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Answer:

5.0 ft-lbf

Step-by-step explanation:

The force is

$F = \dfrac{9}{6^x}$

This force is not a constant force. For a non-constant force, the work done, W, is

$W = \int\limits^{x_2}_{x_1} {F(x)} \, dx$

with $x_1$ and $x_2$ the initial and final displacements respectively.

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3 years ago

I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).

Aleksandr-060686 [28]

Answer:

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

Step-by-step explanation:

$\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$

Let's first isolate the trig function.

Add 1 one on both sides:

$\sqrt{3} \tan(x-\frac{\pi}{8})=1$

Divide both sides by $\sqrt{3}$ :

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

Now recall $\tan(u)=\frac{\sin(u)}{\cos(u)}$ .

$\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$

$\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}$

The first ratio I have can be found using $\frac{\pi}{6}$ in the first rotation of the unit circle.

The second ratio I have can be found using $\frac{7\pi}{6}$ you can see this is on the same line as the $\frac{\pi}{6}$ so you could write $\frac{7\pi}{6}$ as $\frac{\pi}{6}+\pi$ .

So this means the following:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

is true when $x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

where $n$ is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

$x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

Add $\frac{\pi}{8}$ on both sides:

$x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi$

Find common denominator between the first two terms on the right.

That is 24.

$x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi$

$x=\frac{7\pi}{24}+n \pi$ (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval $[0,2\pi)$ .

So if $\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$ and we let $u=x-\frac{\pi}{8}$ , then solving for $x$ gives us:

$u+\frac{\pi}{8}=x$ ( I just added $\frac{\pi}{8}$ on both sides.)

So recall $0\le x$ .

Then $0 \le u+\frac{\pi}{8}$ .

Subtract $\frac{\pi}{8}$ on both sides:

$-\frac{\pi}{8}\le u$

Simplify:

$-\frac{\pi}{8}\le u$

So we want to find solutions to:

$\tan(u)=\frac{1}{\sqrt{3}}$ with the condition:

$-\frac{\pi}{8}\le u$

That's just at $\frac{\pi}{6}$ and $\frac{7\pi}{6}$

So now adding $\frac{\pi}{8}$ to both gives us the solutions to:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$ in the interval:

$0\le x$ .

The solutions we are looking for are:

$\frac{\pi}{6}+\frac{\pi}{8}$ and $\frac{7\pi}{6}+\frac{\pi}{8}$

Let's simplifying:

$(\frac{1}{6}+\frac{1}{8})\pi$ and $(\frac{7}{6}+\frac{1}{8})\pi$

$\frac{7}{24}\pi$ and $\frac{31}{24}\pi$

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

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3 years ago

What is the total floor area in inches of a rectangular floor 8ft. by 6ft.?

Anestetic [448]

Answer:the answeer is

= 72 ft³

Step-by-step explanation:

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