Helppp meeeeeee asap with question 3 pls, arigatoo ;-; y can't i understandd

Mathematics

1 answer:

iris [78.8K]3 years ago

5 0

The answer is B for problem three.

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Step-by-step explanation:

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Troyanec [42]

The solution for a system of equations will be when the graphs touch each other, or intersect, and that happens when both equations equate each other, so, let's do so,

$\bf \begin{cases}
\boxed{y}=3x+10\\
-------\\
x^2+y^2=10\\
y^2=10-x^2\\
y=\sqrt{10-x^2}
\end{cases}\qquad \qquad \boxed{3x+10}=\sqrt{10-x^2}
\\\\\\
\textit{now we square both sides}\qquad (3x+10)^2=(\sqrt{10-x^2})^2
\\\\\\
(3x+10)^2=10-x^2\implies 9x^2+60x+100=10-x^2
\\\\\\
10x^2+60x-90=0\implies x^2+6x-9=0\implies (x+3)(x+3)=0
\\\\\\
x=
\begin{cases}
-3\\
-3
\end{cases}\qquad thus\qquad \boxed{x=-3}$

$\bf -------------------------------\\\\
\textit{now, when x = -3, what is \underline{y}?}\qquad y=3x+10\implies y=3(-3)+10
\\\\\\
y=-9+10\implies \boxed{y=1}\qquad \qquad therefore\qquad (\stackrel{x}{-3}~,~\stackrel{y}{1})$

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3 years ago