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Mashutka [201]
3 years ago
11

Multiply:(square root 10 + 2 square root 8)( square root 10 - 2 square root 8)

Mathematics
2 answers:
kogti [31]3 years ago
7 0
Answer: -22.

Explaination: if you multiply the square root 10+. And Square the root of 8
pantera1 [17]3 years ago
3 0

Answer:

-22

Step-by-step explanation:

We need to multiply the following:

(\sqrt{10} +2\sqrt{8})(\sqrt{10}-2\sqrt{8})

First, we need to FOIL out this expression, then we can simplify it

(\sqrt{10} +2\sqrt{8})(\sqrt{10}-2\sqrt{8})\\\\(\sqrt{10})( \sqrt{10} )+(2\sqrt{8})(\sqrt{10})-(2\sqrt{8})( \sqrt{10})-(2\sqrt{8})(2\sqrt{8}) \\\\10+2\sqrt{80} -2\sqrt{80} -4\sqrt{64} \\\\10-4(8)\\\\10-32\\\\-22

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Answer:

A. 34°

General Formulas and Concepts:

<u>Pre-Algebra</u>

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  • Left to Right

Equality Properties

<u>Trigonometry</u>

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  • [Right Triangles Only] tanθ = opposite over adjacent
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Step-by-step explanation:

<u>Step 1: Define</u>

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<u>Step 2: Solve for </u><em><u>x</u></em>

  1. Substitute in variables [Tangent]:                                                                    \displaystyle tanx^\circ = \frac{24}{35}
  2. Inverse Trig [Tangent]:                                                                                     \displaystyle x^\circ = tan^{-1}(\frac{24}{35})
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A variable force of 3x−2 pounds moves an object along a straight line when it is x feet from the origin. Calculate the work done
Verdich [7]

Answer:

2.70J.

Step-by-step explanation:

We have been given that a variable force of 3x^{-2} pounds moves an object along a straight line when it is x feet from the origin.  We are asked to find the work done in moving the object from x = 1 ft to x = 10 ft.

We know that dW=Fdx, where F represents force.

dW=3x^{-2}dx

Now, we will integrate both sides of equation as:

\int\limits{dw} =\int\limits^{10}_1 {3x^{-2} \, dx

Take the constant out:

\int\limits{1dw} =3\int\limits^{10}_1 {x^{-2} \, dx

w =3\int\limits^{10}_1 {x^{-2} \, dx

Upon applying power rule of integrals, we will get:

w =3\/[ \frac{x^{-2+1}}{-2+1} ]^{10}_1

w =3[ \frac{x^{-1}}{-1} ]^{10}_1

w =3[ -\frac{1}{x} ]^{10}_1

w =3[ -\frac{1}{10} -(-\frac{1}{1})]

w =3[ -\frac{1}{10}+\frac{10}{10})]

w =3[\frac{9}{10}]

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Therefore, the work done by the object is 2.70J.

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