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valentinak56 [21]
2 years ago
8

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

levation to the top of the dam from a point 155 feet farther down river is 0.9 radi- ans; assume the two angle measurements are taken at the same elevation above sea level. How high is the dam?
Mathematics
1 answer:
Nana76 [90]2 years ago
5 0

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 :

Tan θ = opposite / Adjacent

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

x = 148.87307

Using :

h = x Tan 68.755

h = 148.87307 * tan(68.755)

h = 382.92539

h = 382.925 feets

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300 miles

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Answer:

Step-by-step explanation:

Change 2/5 to a decimal

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0.4*5 = 1.133333*x

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2 years ago
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y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\

\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\

This shows that y' = y is true when y = e^x

-----------------------

  • Note 1: A more general solution is y = Ce^x for some constant C.
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Write as an expression: a number that is equal to twice as much as x
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Let's call the number we're looking for y. We know that y is twice as much as x.


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Answer:

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Step-by-step explanation:

12+[(15-5)]+(9-3)]

PEMDAS

Parentheses first

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Then add

12+10+6

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