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stiv31 [10]
3 years ago
8

The top of blue mountain ski slope 19 3/4 yards above sea level. The lowest point of the ski point is 12.5 yards below sea level

. Deangelo and Kashawn are going skiing and will be taking a ski lift up to the top of the mountain. The ski lift is going to pick them up at the midpoint between top and bottom of the slope. At what elevation will they be picked up?
Mathematics
1 answer:
lorasvet [3.4K]3 years ago
5 0

Answer: 3.625 yards above the sea level.


Step-by-step explanation:

Given:  The elevation of top of top of blue mountain above the sea level= 19\frac{3}{4}=19.75\ yards

The length of the lowest point below the sea level= 12.5 yards

The total elevation from the lowest point under the sea and the top  of blue mountain=19.75+12.5=32.25 yards

The midpoint between the top and bottom of the slope = \frac{1}{2}\times32.25=16.125\ yards

Since the ski lift is going to pick them up at the midpoint between top and bottom of the slope.

Therefore, the elevation at which they will be picked up= 19.75-16.125=3.625\ yards above the sea level.

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Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into
Alex_Xolod [135]

Answer:

a) the length of the wire for the circle = (\frac{60\pi }{\pi+4}) in

b)the length of the wire for the square = (\frac{240}{\pi+4}) in

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

         b=\frac{y}{4}

Area of the square which is L² can now be said to be;

A_S=(\frac{y}{4})^2 = \frac{y^2}{16}

On the otherhand; let the radius (r) of the  circle be;

2πr = 60-y

r = \frac{60-y}{2\pi }

Area of the circle which is πr² can now be;

A_C= \pi (\frac{60-y}{2\pi } )^2

     =( \frac{60-y}{4\pi } )^2

Total Area (A);

A = A_S+A_C

   = \frac{y^2}{16} +(\frac{60-y}{4\pi } )^2

For the smallest possible area; \frac{dA}{dy}=0

∴ \frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0

If we divide through with (2) and each entity move to the opposite side; we have:

\frac{y}{18}=\frac{(60-y)}{2\pi}

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

y= \frac{240}{\pi+4}

∴ since y= \frac{240}{\pi+4}, we can determine for the length of the circle ;

60-y can now be;

= 60-\frac{240}{\pi+4}

= \frac{(\pi+4)*60-240}{\pi+40}

= \frac{60\pi+240-240}{\pi+4}

= (\frac{60\pi}{\pi+4})in

also, the length of wire for the square  (y) ; y= (\frac{240}{\pi+4})in

The smallest possible area (A) = \frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

4 0
3 years ago
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