Solve the system of equations by elimination. HELP ASAP!!

Mathematics

1 answer:

ivann1987 [24]4 years ago

7 0

Answer:

{x,y,z} = {1,-1,-2}

Step-by-step explanation:

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Rewrite 9cos 4x in terms of cos x.

rosijanka [135]

$\bf \qquad \textit{Quad identities}\\\\
sin(4\theta )=
\begin{cases}
8sin(\theta )cos^3(\theta )-4sin(\theta )cos(\theta )\\
4sin(\theta )cos(\theta )-8sin^3(\theta )cos(\theta )
\end{cases}
\\\\\\
cos(4\theta)=8cos^4(\theta )-8cos^2(\theta )+1\\\\
-------------------------------\\\\
9cos(4x)\implies 9[8cos^4(x)-8cos^2(x)+1]
\\\\\\
72cos^4(x)-72cos^2(x)+9$

---------------------------------------------------------------------------

as far as the previous one on the 2tan(3x)

$\bf tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\qquad tan({{ \alpha}} + {{ \beta}}) = \cfrac{tan({{ \alpha}})+ tan({{ \beta}})}{1- tan({{ \alpha}})tan({{ \beta}})}\\\\
-------------------------------\\\\$

$\bf 2tan(3x)\implies 2tan(2x+x)\implies 2\left[ \cfrac{tan(2x)+tan(x)}{1-tan(2x)tan(x)}\right]
\\\\\\
2\left[ \cfrac{\frac{2tan(x)}{1-tan^2(x)}+tan(x)}{1-\frac{2tan(x)}{1-tan^2(x)}tan(x)}\right]\implies 2\left[ \cfrac{\frac{2tan(x)+tan(x)-tan^3(x)}{1-tan^2(x)}}{\frac{1-tan(x)-2tan^3(x)}{1-tan^2(x)}} \right]
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$\bf 2\left[ \cfrac{2tan(x)+tan(x)-tan^3(x)}{1-tan^2(x)}\cdot \cfrac{1-tan^2(x)}{1-tan(x)-2tan^3(x)} \right]
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2\left[ \cfrac{3tan(x)-tan^3(x)}{1-tan^2(x)-2tan^3(x)} \right]\implies \cfrac{6tan(x)-2tan^3(x)}{1-tan^2(x)-2tan^3(x)}$

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3 years ago

For Bill's birthday his mom is bringing donuts to school.

kipiarov [429]

36 donuts (2 1/2 dozen) = $8.00

So, you divide 8 by 36 and you get about 22¢ for each donut. Then, you do .22 x 12 which equals $2.64.

One dozen donuts = $2.64

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3 years ago

Solve: 4-(9g-5)=-27 please help

8_murik_8 [283]

Answer:

-2

Step-by-step explanation:

9g-4-5=-27

9g-9=-27

Add 9 to both sides

9g=-18

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3 years ago

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prohojiy [21]

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Step-by-step explanation:

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3 years ago

Solve the equation 12+ 0.35×=20.05

lbvjy [14]

12 + 0.35X = 20.05
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8 0

3 years ago