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.

$\textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases} \\\\[-0.35em] ~\dotfill$

$\begin{cases} h=0\\ k=0\\ a=2\\ c=4 \end{cases}\implies \cfrac{(x-0)^2}{2^2}-\cfrac{(y-0)^2}{b^2} \\\\\\ c^2=a^2+b^2\implies 4^2=2^2+b^2\implies 16=4+b^2\implies \underline{12=b^2} \\\\\\ \cfrac{(x-0)^2}{2^2}-\cfrac{(y-0)^2}{12}\implies \boxed{\cfrac{x^2}{4}-\cfrac{y^2}{12}}$

5 0

2 years ago

QUESTION 1

lukranit [14]

1.) 2368000 (two million, three hundred sixty-eight thousand)
2.) 864,346,000 rounded to nearest thousand
3.) 864,345,537.00 not 100% on the last answer

Good luck

8 0

3 years ago

What is the volume of the sphere?

Eddi Din [679]

V = 4/3 * pi * r^3
V= 4/3 3.14 * 3^3
V = 4/3 *3.14 * 27
27 * 3.14 = 84.78
V = 4/3 * 84.78 = 113.04
V = 113.04
hope this helps

7 0

3 years ago

Read 2 more answers

Find the value of x which satisfies the following equation. log2(x−1)+log2(x+5)=4

weqwewe [10]

$\quad \huge \quad \quad \boxed{ \tt \:Answer }$