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3 years ago

Help please !! I’m am struggling with this lol

Mathematics

1 answer:

tatyana61 [14]3 years ago

4 0

Answer:

x ≈ 3.9

Step-by-step explanation:

Call the segment shared by the two triangle "y". The Pythagorean theorem tells you ...

sum of squares of sides = square of hypotenuse

y² +6² = 10²

x² +7² = y²

Substituting for y² using the second equation, we get ...

x² +7² +6² = 10²

x² = 10² -7² -6² = 100 -49 -36 = 15

Taking the square root, we find ...

x = √15 ≈ 3.9

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Solve: 62x - 3 = 6-2x+1

MatroZZZ [7]

Answer:

x = 5/32 = 0.156

Step-by-step explanation:

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3 years ago

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Please help me find the value of x the answer must be exact

jenyasd209 [6]

Answer:

$x=11$

Step-by-step explanation:

This is a 45°-45°-90° triangle. The two legs are the same length, and the hypotenuse is that length times the square root of two:

$leg=x\\hypotenuse=x\sqrt{2}$

Therefore, the value of x is 11.

You can double-check using the sine ratio:

$sineX=\frac{opposite}{hypotenuse}$

Insert the values:

$sin45=\frac{x}{11\sqrt{2} }$

Insert the equation into a calculator:

$x=11$

:Done

*It's important to know that, when working with trigonometry ratios, the hypotenuse is never considered the adjacent or opposite side, and the 90° angle is never used in the ratios.

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3 years ago

Find the area of the region that lies inside the first curve and outside the second curve.

marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = $\dfrac{1}{2}$

$\theta = -\dfrac{\pi}{3}, \ \ \dfrac{\pi}{3}$

Now, the area of the region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

$A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \ \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ 5^2 d \theta$

$A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \ \theta d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ d \theta$

$A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} \dfrac{cos \ 2 \theta +1}{2} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}$

$A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} {cos \ 2 \theta +1} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}$

$A =25 \begin {bmatrix} \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}} \ \ - \dfrac{25}{2} \begin {bmatrix} \dfrac{2 \pi}{3} \end {bmatrix}$

$A =25 \begin {bmatrix} \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3}) \end {bmatrix} - \dfrac{25 \pi}{3}$

$A = 25 \begin{bmatrix} \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} + \dfrac{\pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}$

$A = 25 \begin{bmatrix} \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}$

$A = \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}$

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx

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3 years ago

1. William earns $134,644.80 per year, and has his federal taxes withheld from each weekly check. In addition, a total of $53.60

SashulF [63]

I only have the answer for 3 and i believe it is 20,090

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3 years ago

Witch is greater -4 or 8/2

OLga [1]

8/2 is the greater one.

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