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Schach [20]
2 years ago
15

The picture shows a box sliding down a ramp.

Mathematics
2 answers:
forsale [732]2 years ago
8 0
In your question where as the picture shows a box sliding down a ramp and ask to calculate the distance, in feet, that the box has to travel to move form point A to point C and the best answer would be 12 cosec 65. I hope you are satistfied with my answer and if you need some clarification, please feel free to ask for more
bija089 [108]2 years ago
6 0

Answer:

AC=12cosec(65\°)\ ft

Step-by-step explanation:

we know that

In the right triangle ABC

sin(65\°)=\frac{AB}{AC}

In this problem we have

AB=12\ ft

Substitute and solve for AC

sin(65\°)=\frac{12}{AC}

AC=\frac{12}{sin(65\°)}

Remember that

cosec(65\°)=\frac{1}{sin(65\°)}

substitute

AC=12cosec(65\°)\ ft

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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
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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)

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Answer:

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Answer:

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Answer:

A and B are independent events.

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