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3 years ago

HELP PLEASE !!!!!

Mathematics

1 answer:

Ivahew [28]3 years ago

3 0

$\sum_{k=4}^8(2k+2)=2\cdot4+2+2\cdot5+2+2\cdot6+2+2\cdot7+2+2\cdot8+2\\
\sum_{k=4}^8(2k+2)=10+12+14+16+18\\
\sum_{k=4}^8(2k+2)=70$

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2 years ago

The top of the Boulder Dam has an angle of elevation of 1.2 radians from a point on the Colorado River. Measuring the angle of e

Nana76 [90]

Answer:

382.925 feets

Step-by-step explanation:

The solution diagram is attached below :

Converting radian measurement to degree :

radian angle * 180/π = degree angle

1.2 * 180/π = 68.755°

0.9 * 180/π = 51.566°

Height of dam is h:

Using trigonometry :

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Tan 68.755° = h / x

h = x Tan 68.755° - - - (1)

Tan 51.566° = h / (155+x)

h = (155+x) tan 51.566° - - - (2)

Equate (1) and (2)

x Tan 68.755 = (155+x) Tan 51.566

x Tan 68.755 = 155tan 51.566 + x tan 51.566

x Tan 68.755 = 195.32311 + x Tan 51.566

x Tan 68.755 - x Tan 51.566 = 195.32311

x(tan 68.755 - tan 51.566) = 195.32311

x * 1.3120110 = 195.32311

1.3120110x = 195.32311

x = 195.32311 / 1.3120110

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Using :

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3 years ago

If Logx (1 / 8) = - 3 / 2, then x is equal to A. - 4 B. 4 C. 1 / 4 D. 10

Basile [38]

The answer to this is b

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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

7 0

3 years ago