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Alisiya [41]
3 years ago
7

The flagpole at school casts a shadow that is 18 feet long. A student nearby is 5.5 feet tall and casts a shadow that is 3 feet

long. How tall is the flagpole?
Mathematics
1 answer:
azamat3 years ago
5 0

Answer:

33 feet.

Step-by-step explanation:

If the ratio of the student to their shadow and flagpole to its shadow are equal, then you can use the proportion

5.5/3 = x/18 and cross multiply to find x.

5.5 × 18 = 99

3 × x = 3x

3x/3 = x, 99/3 = 33.

x = 33.

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Arisa [49]
-7d > 21
d < -3

Remember the sign flips when you multiply or divide by a negative number.
3 0
3 years ago
25to300 round to the nearest percent
mihalych1998 [28]

Answer:

163%

Step-by-step explanation:

Find in-between 25 and 300, which is 162.5 ,round to the nearest percentage being 163%. Hence your answer

8 0
2 years ago
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snow_lady [41]
It's 8! range is just the greatest value minus the lowest value
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5 0
2 years ago
jim had 20 to spend at the video rental he has a late fee of 4.00 and it will cost him 2.99 to rent each movie not including tax
Anastasy [175]

Answer:

2.99m-4 (less than or equal to symbol after) 20

Step-by-step explanation:

This would come out too 5 movies because 20-4= 16 16-2.99=5.3 but it is impossible to buy 5.3 movies so you would round it down to 5

8 0
3 years ago
A ball is thrown downward from a cliff. Its position at time t seconds is given by the formula s(t) = 16t2 + 32t, where s is in
Rom4ik [11]

Answer:

It takes 2.8 seconds for the ball to fall 215 ft.

Step-by-step explanation:

We are given a position function s(t) where s stands for the number of feet the ball has fallen, so we have to replace s with the given value of 215 ft and solve for the time t.

Setting up the equation.

The motion equation is given by

s(t) =16t^2+32t

We can replace there s = 215 ft to get

215=16t^2+32t

Solving for the time t.

From the previous equation we can move all terms in one side to get

16t^2+32t-215=0

At this point we can solve for t using quadratic formula.

t = \cfrac{-b\pm \sqrt{b^2-4ac}}{2a}

where a, b and c are the coefficients of the quadratic equation

at^2+bt+c=0

So we get

a=16\\b=32\\c=-215

Replacing on the quadratic formula we get

t = \cfrac{-32\pm \sqrt{32^2-4(16)(-215)}}{2(16)}

Using a calculator we get

t=-4.8 , t = 2.8

Physically speaking the only result that makes sense is to move forward in time that give us t = 2.8 seconds.

We can conclude that it takes 2.8 seconds for the ball to fall 215 ft.

3 0
3 years ago
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