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Lesechka [4]
2 years ago
12

X to the third plus x minus 7 equals -3 square root of x - 1

Mathematics
1 answer:
Ugo [173]2 years ago
5 0

Answer:

  x ≈ 1.5004

Step-by-step explanation:

We suppose your equation is ...

  x^3+x-7=-3\sqrt{x-1}

Squaring both sides and subtracting the right side gives ...

  x^6 +2x^4 -14x^3 +x^2 -23x +58 = 0

This 6th-degree equation has two positive real roots, near x=1.5, and x=2. The root at x=2 is extraneous. The one near x=1.5 is irrational.

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Use the triangle below to find cos V.
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Answer:

cos\ V = \frac{2\sqrt{29}}{29}

Step-by-step explanation:

Given

The attached triangle

Required

Find cos V

In trigonometry:

cos\theta = \frac{Adjacent}{Hypotenuse}

In this case:

cos V= \frac{TV}{UV}

Where

TV = 6

and

UV^2 = TV^2 + TU^2 -- Pythagoras

UV^2 = 6^2 + 15^2

UV^2 = 261

Take square roots

UV = \sqrt{261

UV = \sqrt{9*29

UV = \sqrt{9} *\sqrt{29

UV = 3\sqrt{29

So:

cos V= \frac{TV}{UV}

cos\ V = \frac{6}{3\sqrt{29}}

cos\ V = \frac{2}{\sqrt{29}}

Rationalize:

cos\ V = \frac{2}{\sqrt{29}}*\frac{\sqrt{29}}{\sqrt{29}}

cos\ V = \frac{2\sqrt{29}}{29}

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