Xˆ8 ÷ xˆ2 fully factorised

Mathematics

2 answers:

Ksivusya [100]3 years ago

3 0

Answer:

Step-by-step explanation:

x^8/x^2= $x^{8-2}$ = x^6

when you divide same base and different powers, you subtract the powers

Yuki888 [10]3 years ago

3 0

Answer:

Step-by-step explanation:

There are 2 ways that you can do this problem. The first is just to factor everything and cancel.

x * x * x * x *x * x *x *x

====================

x * x

Two of the xs on the top cancel leaving you with the number 1 on the bottom

x * x * x * x * x * x = x^6

This is one of your answers.

The other is you can notice you get the same thing by subtracting powers.

x^(8 - 2) = x^6

I'm not sure which answer is expected. I would guess the second one.

You might be interested in

Brent would like to start a small landscape business this summer. He has a lawn mower, but needs a truck to carry the lawn mower

spin [16.1K]

Answer:

I would buy the 2000 Toyota Tundra. Yes it is the most expensive, but it is the newest so it should require less expense money for problems, it is a manual transmission 6 cylinder so it should get better gas mileage, and it has fewer miles. The rankings of these trucks are not provided, however a Toyota tundra is considered to be a workhorse in the truck. If this is an opinion question, that's my choice.

Step-by-step explanation:

3 0

3 years ago

Sin^2(x/2)=sin^2x <br> Find the value of x

Veronika [31]

$\sin^2\dfrac x2=\sin^2x$
$\dfrac{1-\cos x}2=1-\cos^2x$
$1-\cos x=2-2\cos^2x$
$2\cos^2x-\cos x-1=0$
$(2\cos x+1)(\cos x-1)=0$
$\implies\begin{cases}2\cos x+1=0\\\cos x-1=0\end{cases}$ $\implies\begin{cases}\cos x=-\frac12\\\cos x=1\end{cases}$

The first case occurs in $0\le x$ for $x=\dfrac{2\pi}3$ and $x=\dfrac{4\pi}3$ . Extending the domain to account for all real $x$ , we have this happening for $x=\dfrac{2\pi}3+2n\pi$ and $\dfrac{4\pi}3+2n\pi$ , where $n\in\mathbb Z$ .

The second case occurs in $0\le x$ when $x=0$ , and extending to all reals we have $x=2n\pi$ for $n\in\mathbb Z$ , i.e. any even multiple of $\pi$ .