whats the solution to these inequalities??

[vlh]

x^2+(y+2)^2<16
(x+1)^2/64+(y+2)^2>1

 

[Deleted member 4993]

x^2+(y+2)^2<16
(x+1)^2/64+(y+2)^2>1

Please show us your work, indicating exactly where you are stuck - so that we may know where to begin to help you.

 

[vlh]

I don't even know how to start it

 

[BigGlenntheHeavy]

$\displaystyle Here \ is \ the \ graphs \ of \ x^2+(y+2)^2 \ = \ 16 \ and \ \frac{(x+1)^2}{8^2}+\frac{(y+2)^2}{1^2} \ = \ 1.$

[attachment=0:3vwrv8uo]pqr.jpg[/attachment:3vwrv8uo]

 

[soroban]

Hello, vlh!

Here's a start . . .

\(\displaystyle \begin{array}{cccc}x^2+(y+2)^2 &<& 16 & [1] \\ \\[-3mm]\dfrac{(x+1)^2}{64}+(y+2)^2 &>& 1 & [2] \end{array}\)

$\displaystyle \begin{array}{ccccccc}\text{Multiply [2] by -1:} & -\dfrac{(x+1)^2}{64} - (y+2)^2) &<& -1 \\ \\ \text{Add [1]:} & x^2 + (y+2)^2 &<& 16 \\ \\ \text{And we have:} & x^2 - \dfrac{(x+1)^2}{64} & <& 15 \\ \\ \text{which simplifies: } & 63x^2 - 2x - 961 &<& 0 \end{array}$


$\displaystyle \text{This is an up-opening parabola. }\;\text{It is negative between its x-intercepts.}$


$\displaystyle \text{Quadratic Formula }\;x \;=\;\frac{2 \pm\sqrt{242176}}{126} \;\approx\;\begin{Bmatrix}3.922 \\ \text{-}3.890\end{Bmatrix}$


$\displaystyle \text{Therefore: }\;-3.890 \;<\;x\;<\;3.922$

. . which seems to agree with BigGlenn's graph
.