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1 year ago

to solve the system of equations below, Becca isolated x^2 in the firs equation and then substituted it into the second equation

1 answer:

4 0

The resulting equation if Becca isolated x² in the first equation and then substituted it into the second equation is (9-y²) / 25 - y²/36 = 1

<h3>Equation</h3>

x² + y² = 9

x²/25 - y²/36 = 1

From (1)

x² = 9 - y²

substitute x² = 9 - y² into (2)

x²/25 - y²/36 = 1

(9-y²) / 25 - y²/36 = 1

Therefore, the resulting equation is option C; (9-y²) / 25 - y²/36 = 1

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3 years ago

Hi. Please I need help with these questions.

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Q 12 roots of the equation

$2x^{2} -6+1=0 \\= (x-\frac{\sqrt{10} }{2})(x+\frac{\sqrt{10} }{2})\\$

∝ = $\frac{\sqrt{10} }{2}$

β = $-\frac{\sqrt{10} }{2}$

no matter if u oppose the root

(i) 2( $\frac{\sqrt{10} }{2}$ ) $(-\frac{\sqrt{10} }{2} )^{2}$ +2 $(\frac{\sqrt{10} }{2} )^{2}$ ( $-\frac{\sqrt{10} }{2}$ )+2(

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Q 13 roots of equation

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the equation is

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3 years ago

PRECALC QUESTION! Convert the polar representation of this complex number into its standard form: z = 2(cos 11pi/6+ i sin 11pi/6

Tanzania [10]

$\bf z=2\left[ cos\left( \frac{11\pi }{6} \right) + i\ sin\left( \frac{11\pi }{6} \right) \right]\qquad 
\begin{cases}
r=2\\
\theta=\frac{11\pi }{6}\\
-----\\
a=x=rcos(\theta)\\
b=y=rsin(\theta)
\end{cases}\implies a+bi
\\\\\\
2cos\left( \frac{11\pi }{6} \right)+2sin\left( \frac{11\pi }{6} \right)\ i\implies 2\cdot \cfrac{\sqrt{3}}{2}+2\cdot \cfrac{-1}{2}\ i\implies \sqrt{3}-1i
\\\\\\
\boxed{\sqrt{3}-i}$

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3 years ago

Read 2 more answers

Line joins ordered pairs negative 4, 3, 2, negative 3. Second joins negative 4, negative 3, 2, 3. A shaded above first second li

xxMikexx [17]

Answer

That should be it!

Step-by-step explanation:

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3 years ago

Read 2 more answers

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marin [14]

Answer:

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3 years ago