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3 years ago

5

Please , I didn't have time to study. god bless to all <3

1 answer:

Naily [24]3 years ago

3 0

This is the first half of the question.
Let's solve your equation step-by-step.5x=9x3+2x2−5x+4Step 1: Subtract 9x^3+2x^2-5x+4 from both sides.5x−(9x3+2x2−5x+4)=9x3+2x2−5x+4−(9x3+2x2−5x+4)−9x3−2x2+10x−4=0Step 2: Use cubic formula.x=−1.31953Answer:<span>x=<span>−1.31953

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Answer the question in the picture

nekit [7.7K]

Recall the angle sum identities:

$\sin(x+y)=\sin x\cos y+\cos x\sin y$

$\cos(x+y)=\cos x\cos y-\sin x\sin y$

Now,

$\tan(x+y)=\dfrac{\sin(x+y)}{\cos(x+y)}=\dfrac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}$

Divide through numerator and denominator by $\cos x\cos y$ to get

$\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$

Next, we use the fact that $x,y$ lie in the first quadrant to determine that

$\sin x=\dfrac12\implies\cos x=\sqrt{1-\sin^2x}=\dfrac{\sqrt3}2$

$\cos y=\dfrac{\sqrt2}2\implies\sin x=\sqrt{1-\cos^2x}=\dfrac1{\sqrt2}$

So we then have

$\tan x=\dfrac{\sin x}{\cos x}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}$

$\tan y=\dfrac{\sin y}{\cos y}=\dfrac{\frac1{\sqrt2}}{\frac{\sqrt2}2}=1$

Finally,

$\tan(x+y)=\dfrac{\frac1{\sqrt3}+1}{1-\frac1{\sqrt3}}=\dfrac{1+\sqrt3}{\sqrt3-1}=2+\sqrt3\approx3.73$

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Step-by-step explanation:

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172 = 1/2 * x + 45
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3 years ago

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maks197457 [2]

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Step-by-step explanation:

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