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jeyben [28]
3 years ago
5

X^2-4x+40=0 what is the roots

Mathematics
2 answers:
lys-0071 [83]3 years ago
8 0
I think you can't find the root for this equation...
trasher [3.6K]3 years ago
3 0
X^2-4x=0-40
2x-4x=-40
-2x=-40
x=0.05
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3 years ago
3n+75=50+2n please help righting more
alexandr1967 [171]

Answer: n = -25

Step-by-step explanation:

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5 0
2 years ago
If cos(θ)=2853 with θin Q IV, what is sin(θ)?
marysya [2.9K]

Answer: \sin \theta=\frac{-45}{53}

Step-by-step explanation:

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\cos\theta=\frac{28}{53}

And we know that θ is in the Fourth Quadrant.

So, Except cosθ and sec θ, all trigonometric ratios will be negative.

As we know the "Trigonometric Identity":

\cos^2\theta+\sin^2\theta=1\\\\\sin \theta=\sqrt{1-\cos^2\theta}\\\\\sin \theta=\sqrt{1-(\frac{28}{53})^2}=\sqrt{\frac{53^2-28^2}{53^2}}\\\\\sin \theta=\sqrt{\frac{2025}{53^2}}\\\\\sin \theta=\frac{45}{53}

It must be negative due to its presence in Fourth quadrant.

Hence, \sin \theta=\frac{-45}{53}

7 0
3 years ago
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