Prime factorization of 2·5·2=2^2·5
1. Observe problem
Solve for the area and Circumference of a circle who's radius is 3 cm
Area for circle formula:
pi*r^2=
9pi
I simplicity is needed, 9 pi is around
28.2743338823cm^2
Circumference for circle formula: 2*pi*r
which is
6pi
if simplicity is again needed, its around
18.8495559215cm
Check the picture below, so the hyperbola looks more or less like so, so let's find the length of the conjugate axis, or namely let's find the "b" component.
![\textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Ctextit%7Bhyperbolas%2C%20horizontal%20traverse%20axis%20%7D%20%5C%5C%5C%5C%20%5Ccfrac%7B%28x-%20h%29%5E2%7D%7B%20a%5E2%7D-%5Ccfrac%7B%28y-%20k%29%5E2%7D%7B%20b%5E2%7D%3D1%20%5Cqquad%20%5Cbegin%7Bcases%7D%20center%5C%20%28%20h%2C%20k%29%5C%5C%20vertices%5C%20%28%20h%5Cpm%20a%2C%20k%29%5C%5C%20c%3D%5Ctextit%7Bdistance%20from%7D%5C%5C%20%5Cqquad%20%5Ctextit%7Bcenter%20to%20foci%7D%5C%5C%20%5Cqquad%20%5Csqrt%7B%20a%20%5E2%20%2B%20b%20%5E2%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)

Answer:
It is not equivalent.
Step-by-step explanation:
You have to expand the brackets :
(x-4)(x+3)
= x(x) + x(3) - 4(x) - 4(3)
= x² + 3x - 4x - 12
= x² - x - 12
Therefore, x² - x - 12 / (x-4)(x+3) is not equivalent to x² - 4x - 12.
Answer:
18
It’s literally 72 divided by 4 which equals 18