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2 years ago

Help me please show how you got the answer

Mathematics

1 answer:

Lilit [14]2 years ago

3 0

Answer:

Firstly you need to put the equation in form of. y =ax+b

Let us do it

We have

2x-2y=18

-2y=18-2x

-y=(18) /2 - (2x)/2

-y =9-x

Mutiply by - to remove it beside y

Y=-9+x

Which is y=x-9

Slope =a=1

Yintercept=b=-9

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3 years ago

Describe the figure attached using correct geometric names<br> perpendicular, straight, etc.

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3 years ago

Rewrite 9cos 4x in terms of cos x.

rosijanka [135]

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

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

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

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

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

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

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3 years ago

How do you solve +5=-5a+5

Bogdan [553]

You subtract the number (5) on the other side of the = sign from the number on the side of the = sign. and divide (-5a) by the number you subtracted before. So untimely- 5-5= 0 ÷ -5a = 0

6 0

3 years ago