Could some help me thanks

Mathematics

1 answer:

dezoksy [38]3 years ago

3 0

You are given 2/3 angles to solve for x;

24+56 = 90

90 + x = 180

x = 90 degrees

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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)

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

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

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

4 0

3 years ago

Two-thirds of a number added to itself is 20. What is the number?

vivado [14]

Answer:

Step-by-step explanation:

2/3 x + x = 20

2/3 x + 3/3x = 20

5/3 x = 20 ... times 3/5 both sides

x = 12

check: 2/3 x 12 + 12 = 8 + 12 = 20

8 0

3 years ago

Help on college algebra, question is on the picture!

kodGreya [7K]

$\frac{7x+1}{x-2} + \frac{2}{x} = \frac{-4}{ x^{2} -2x}$ is given
here , $\frac{7x+1}{x-2} + \frac{2}{x} = \frac{-4}{ x^{2} -2x}$
or, $\frac{7x^{2} +x +2x-4 }{x(x-2)} = \frac{-4}{ x^{2} -2x}$
or, $\frac{7 x^{2} +3x-4}{ x^{2} -2x} = \frac{-4}{ x^{2} -2x}$
or,7x²+3x-4=-4
or,7x²+3x=0
or,x(7x+3)=0
Either,x=0| Or,7x+3=0
|or,7x=-3
|or,x= $\frac{-3}{7}$
So , the solution set is S={0,- $\frac{3}{7}$ }