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3 years ago

Graph the function. [-X + 1, -10 6. f(x) = x2-9, -3 (2x+1, 3

Mathematics

1 answer:

lapo4ka [179]3 years ago

6 0

Answer:

2x2 - 5x - 12 = 0.

(2x + 3)(x - 4) = 0.

2x + 3 = 0 or x - 4 = 0.

x = -3/2, or x = 4.

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Korvikt [17]

Each band member would get 4 tickets.
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116/29=4

8 0

3 years ago

How do i solve this question

frozen [14]

$\bf \begin{cases}
h(x)=(f\circ g)(x)\\
h(x)=\sqrt{x+5}\\
f(x)=\sqrt{x+2}
\end{cases}\\\\
-------------------------------\\\\
(f\circ g)(x)\implies f[\ g(x)\ ]\implies f[\ g(x)\ ]=\sqrt{g(x)+2}\qquad thus
\\\\\\
\sqrt{g(x)+2}=(f\circ g)(x)=h(x)=\sqrt{x+5}\qquad therefore
\\\\\\
\sqrt{g(x)+2}=\sqrt{x+5}\impliedby \textit{squaring both sides}\\\\\\
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3 years ago

This extreme value problem has a solution with both a maximum value and a minimum value. use lagrange multipliers to find the ex

noname [10]

$L(x,y,z,\lambda)=10x+10y+2z+\lambda(5x^2+5y^2+2z^2-42)$

$L_x=10+10\lambda x=0\implies1+\lambda x=0$

$L_y=10+10\lambda y=0\implies1+\lambda y=0$

$L_z=2+4\lambda z=0\implies1+2\lambda z=0$

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$\begin{cases}L_x=0\\L_y=0\\L_z=0\end{cases}\implies1+\lambda x=1+\lambda y=1+2\lambda z=0\implies x=y=2z$

$5x^2+5y^2+2z^2=5(2z)^2+5(2z)^2+2z^2=42z^2=42\implies z^2=1$

$z^2=1\implies z=\pm1\implies x=y=\pm2$

There are two critical points, at which we have

$f\left(2,2,1\right)=42\text{ (a maximum value)}$

$f\left(-2,-2,-1\right)=-42\text{ (a minimum value)}$

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3 years ago

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kherson [118]

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2 years ago

2+2-2-3-87+20090 then split in half

Svetllana [295]

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4 years ago