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Anna [14]
3 years ago
9

Sally has Christmas lights the roof is 16 feet high the ladder is 20 ft long how far away is letter from wall

Mathematics
1 answer:
lesantik [10]3 years ago
4 0
It should be 12 ft away from the wall. Not 4 ft. You do not just subtract the difference because as you pull it away from the wall, the ladder is mostly moving horizontally with only a small vertical change.

You can use Pythagorean Theorem to find the answer. Think of the ladder as the hypotenuse of a right triangle with the wall and the ground as the legs.

a^2 + b^2 = c^2

16^2 + b^2 = 20^2

b^2 = 20^2 - 16^2

b^2 = 144

b = 12 ft






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Answer:

y= -5x +7

Step-by-step explanation:

We can see points on the graph:

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The function in general form is:

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Based on the found values of m and b, the given line is:

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The position of an object along a vertical line is given by s(t) = −t3 + 3t2 + 7t + 4, where s is measured in feet and t is meas
saw5 [17]

Answer:

The maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

Step-by-step explanation:

Given : The position of an object along a vertical line is given by s(t) = -t^3+3t^2+7t +4, where s is measured in feet and t is measured in seconds.

To find : What is the maximum velocity of the object in the time interval [0, 4]?

Solution :

The velocity is rate of change of distance w.r.t time.

Distance in terms of t is given by,

s(t) = -t^3+3t^2+7t +4

Derivate w.r.t. time,

v(t)=s'(t) = -3t^2+6t+7

It is a quadratic function so its maximum is at vertex of the function.

The x point of the function is given by,

x=-\frac{b}{2a}

Where, a=-3, b=6 and c=7

t=-\frac{6}{2(-3)}

t=-\frac{6}{-6}

t=1

As 1 lie between interval [0,4]

Substitute t=1 in the function,

v(t)= -3(1)^2+6(1)+7

v(t)= -3+6+7

v(1)=10

Th maximum velocity is 10 ft/s.

Therefore, the maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

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