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Dahasolnce [82]
2 years ago
7

You and a friend want to know how much rope you need to climb a large rock.  Your friend is 5.5 feet tall and casts a shadow tha

t is 18 feet long.  If the rock casts a shadow that is 209.5 feet long, calculate the height of the rock from the ground.  Round your answer to the nearest foot.  (Hint: make a drawing)

Mathematics
1 answer:
siniylev [52]2 years ago
6 0
The drawing shows the rock height to be 64 ft.

The shadow lengths are proportional to the object heights, so we have
   (rock height)/(209.5 ft) = (5.5 ft)/(18 ft)
   (rock height) = (209.5 ft)*(5.5/18) ≈ 64.01 ft

The height of the rock from the ground is about 64 ft.

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There are two similar right triangles. One has side lengths of 6, 8, and 10. The other has side lengths of 24, and 18. What is t
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Answer:

30

Step-by-step explanation:

24/8 = 3

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10x3 = 30

3 0
3 years ago
Last one plz help me!
Mashutka [201]

Answer:

a) no real solutions

Step-by-step explanation:

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A bracelet that regularly sells for $44 is on sale for 25% off.Find the sale price of the bracelet.
a_sh-v [17]

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$33

Step-by-step explanation:

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3 years ago
Read 2 more answers
A rectangle has a perimeter of 30 ft. Find a function that models its area A in terms of the length x of one of its sides. What
Vedmedyk [2.9K]

Answer:

Step-by-step explanation:

Let the other side of the rectangle be y. The perimeter of the rectangle is expressed as P = 2(x+y)

Given P = 30ft, on substituting P = 30 into the expression;

30 = 2(x+y)

x+y = 15

y = 15-x

Also since the area of the rectangle is xy;

A = xy

Substitute y = 15-x into the area;

A = x(15-x)

A = 15x-x²

The function that models its area A in terms of the length x of one of its sides is A = 15x-x²

The side of length x yields the greatest area when dA/dx = 0

dA/dx = 15-2x

15-2x = 0

-2x = -15

x = -15/-2

x = 7.5 ft

Hence the side length, x that yields the greatest area is 7.5ft.

Since y = 15-x

y = 15-7.5

y = 7.5

Area of the rectangle = 7.5*7.5

Area of the rectangle = 56.25ft²

6 0
2 years ago
How do you find absolute extrema for a function?<br> f(x)= (8+x)/(8-x); Interval of [4,6]
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\bf f(x)=\cfrac{8+x}{8-x}\implies \cfrac{dy}{dx}=\stackrel{quotient~rule}{\cfrac{1(8-x)-(8+x)(-1)}{(8-x)^2}}\implies \cfrac{dy}{dx}=\cfrac{16}{(8-x)^2}

now, we get critical points from zeroing out the derivative, and also from zeroing out the denominator, but those at the denominator are critical points where the function is not differentiable, namely a sharp spike or cusp or an asymptote.

so, from zeroing out the derivative we get no critical points there, from the denominator we get x = 8, but can't use it because f(x) is undefined.

therefore, we settle for the endpoints, 4 and 6,

f(4) =3    and      f(6) = 7

doing a first-derivative test, we see the slope just goes up at both points and in between, but the highest is f(6), so the absolute maximum is there, while we can take say f(4) as the only minimum and therefore the absolute minumum as well.
3 0
3 years ago
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