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Svetllana [295]
2 years ago
10

At a track meet, the longest shot put throw by a boy is 25 feet and 8 inches. The longest shot put throw by a girl is 19 feet an

d 3 inches. How many times greater is the longest shot put throw by a boy than by a girl?
Please explain
Mathematics
2 answers:
MAVERICK [17]2 years ago
4 0
Its 6 feet and 5 inches 
Hunter-Best [27]2 years ago
3 0

Answer:

difference = 6 feet 5 inches

Step-by-step explanation:

The longest shot put throw by a boy is given as 25 feet 8 inches . The longest shot put throw by a girl is given as 19 feet and 3 inches. The number of times the longest shot put throw by a boy is greater than by a girl can be computed as follow.

The question asked us to find the difference between the throw. The difference between the throw is the boy throw minus the girls throw. Mathematically,

The longest boys throw = 25 feet 8 inches

The longest girls throw =  19 feet 3 inches

difference = The longest boy throw  - The longest girl throw

difference =  25 feet 8 inches -   19 feet 3 inches

difference = 6 feet 5 inches

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Solve I = Prt for P when I = 5,480, r = .04, and t = 7.
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Answer: P = 19,571.42857142857

Step-by-step explanation:

Plug in the variables.

5480 = x(.04)(7)

.04*7=.28

5480=.28x

.28x/.28=x

5480/.28=19,571.42857142857

19,571.42857142857(.04)(7)=5480

P = 19,571.42857142857

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2 years ago
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Well, here's one way to do it at least... 

<span>For reference, let 'a' be the side opposite A (segment BC), 'b' be the side opposite B (segment AC) and 'c' be the side opposite C (segment AB). </span>

<span>Let P=(4,0) be the projection of B onto the x-axis. </span>
<span>Let Q=(-3,0) be the projection of C onto the x-axis. </span>

<span>Look at the angle QAC. It has tangent = 5/4 (do you see why?), so angle A is atan(5/4). </span>

<span>Likewise, angle PAB has tangent = 6/3 = 2, so angle PAB is atan(2). </span>

<span>Angle A, then, is 180 - atan(5/4) - atan(2) = 65.225. One down, two to go. </span>

<span>||b|| = sqrt(41) (use Pythagorian Theorum on triangle AQC) </span>
<span>||c|| = sqrt(45) (use Pythagorian Theorum on triangle APB) </span>

<span>Using the Law of Cosines... </span>
<span>||a||^2 = ||b||^2 + ||c||^2 - 2(||b||)(||c||)cos(A) </span>
<span>||a||^2 = 41 + 45 - 2(sqrt(41))(sqrt(45))(.4191) </span>
<span>||a||^2 = 86 - 36 </span>
<span>||a||^2 = 50 </span>
<span>||a|| = sqrt(50) </span>

<span>Now apply the Law of Sines to find the other two angles. </span>

<span>||b|| / sin(B) = ||a|| / sin(A) </span>
<span>sqrt(41) / sin(B) = sqrt(50) / .9080 </span>
<span>(.9080)sqrt(41) / sqrt(50) = sin(B) </span>
<span>.8222 = sin(B) </span>
<span>asin(.8222) = B </span>
<span>55.305 = B </span>

<span>Two down, one to go... </span>

<span>||c|| / sin(C) = ||a|| / sin(A) </span>
<span>sqrt(45) / sin(C) = sqrt(50) / .9080 </span>
<span>(.9080)sqrt(45) / sqrt(50) = sin(C) </span>
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<span>A = 65.225 </span>
<span>B = 55.305 </span>
<span>C = 59.470 </span>
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Answer:

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