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Ludmilka [50]
3 years ago
14

The Kaibab Trail at the Grand Canyon begins at 7,000 feet above sea level. If you descend on the trail 4,500 feet to the Colorad

o River and then hike up 1,375 feet to a campsite, how many feet above sea level are you?
You are____
feet above sea level.
Mathematics
1 answer:
Eddi Din [679]3 years ago
7 0

Answer:

3875 feet above sea level

Step-by-step explanation: First, you would subtract 4,500 from 7,000 because you're going down. Then, since you're going up, you would add 1,375.

7,000- 4,500 = 2,500

2,500+ 1,375 = 3875

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Vedmedyk [2.9K]

Answer: 2605 more on weekdays

Step-by-step explanation:

14239 - 11634 = 2605

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1 year ago
Find the exact value of sin theta/12
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Answer:sin

2

θ

=

−

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169

and

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cos

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Q

4

i.e.

3

π

2

<

θ

<

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(

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13

)

2

−

√

1

−

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√

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Hence,

sin

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sin

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×

(

−

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13

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×

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13

=

−

120

169

and

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Step-by-step explanation:

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3 years ago
Angle relationships with parallel lines please help can someone answer
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x=56

Step-by-step explanation:

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3 years ago
Which of these represents the final step to a no solution problem?
alukav5142 [94]
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3 years ago
Given s left parenthesis t right parenthesis equals 5 t squared plus 5 ts(t)=5t2+5t​, find
finlep [7]

Answer:

The velocity function is v(t)=10t+5.

The acceleration function is a(t)=10.

When t = 44​, the velocity is v(44)=445 \:\frac{ft}{s}.

When t = 44​, the acceleration is a(44)=10\: \frac{ft}{s^2}.

Step-by-step explanation:

We know that the position function is given by

s(t)=5t^2+5t

Velocity is defined as the rate of change of position or the rate of displacement. If you take the derivative of the position function you get the instantaneous velocity function.

v(t)=\frac{ds}{dt}

Acceleration is defined as the rate of change of velocity. If you take the derivative of the instantaneous velocity function you get the instantaneous acceleration function.

a(t)=\frac{dv}{dt}

The instantaneous velocity function is given by

v(t)=\frac{d}{dt} s(t)=\frac{d}{dt}(5t^2+5t)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(5t^2\right)+\frac{d}{dt}\left(5t\right)\\\\\mathrm{Apply\:the\:Power\:Rule}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}\\\\v(t)=10t+\frac{d}{dt}\left(5t\right)\\\\\mathrm{Apply\:the\:common\:derivative}:\quad \frac{d}{dt}\left(t\right)=1\\\\v(t)=10t+5

The instantaneous acceleration function is given by

a(t)=\frac{dv}{dt} =\frac{d}{dt}(10t+5)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\a(t)=\frac{d}{dt}\left(10t\right)+\frac{d}{dt}\left(5\right)\\\\a(t)=10

To find the velocity and acceleration when t = 44, we substitute this value into the velocity and acceleration functions

v(44)=10(44)+5\\v(44)=445 \:\frac{ft}{s}

a(44)=10\: \frac{ft}{s^2}

6 0
4 years ago
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