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

The chemical element einsteinium-

AURORKA [14]

Answer:

Every week, the mass of the sample is multiplied by a factor of 0.81

Step-by-step explanation:

Let's rewrite the base and find the expression whose exponent is just ttt.

(0.97)7t+5=(0.97)7t⋅(0.97)5=(0.977)t⋅(0.97)5

Therefore, we can rewrite the modeling function as follows.

M(t)=(0.97)5⋅(0.977)t

According to this model, the mass of the sample is multiplied by 0.977 every week. Rounding this to two decimal places, we get 0.977≈0.81.

7 0

3 years ago