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Setler [38]
3 years ago
13

A singles tennis court is a rectangle 27 feet wide and 78 feet long. Suppose a player at corner A hits the ball to her opponent

in the diagonally opposite corner B. Approximately how far does the ball travel, to the nearest tenth of a foot?
Mathematics
1 answer:
Sav [38]3 years ago
6 0
Basically, you just have to find the length of the rectangle that is 27 x 78 feet.
The equation for the diagonal: 
d = sqrt(l^2+w^2)
l = 27
w = 78
plug them in and solve
d = sqrt ( (27^2) + (78^2) )
d = sqrt ( 729 + 6084 )
d = sqrt ( 6813 )
d <span>≈ 82.5

The ball traveled approximately 82.5 feet from one corner of the rectangular 27 x 78 foot field, diagonally to the other side.

Hope this helps</span>

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Suppose you throw a dart at a circular target of radius 10 inches. Assuming that you hit the target and that the coordinates of
sukhopar [10]

Answer:

a) The probability is 0.04

b) The probability is 0.36

c) The pprobability is 0,25

d) The probability is 0.09

Step-by-step explanation:

Lets calculate areas:

the target has a radius of 10 inces, hence the target area has a area on 10²*π = 100π square inches.

a) A circle of 2 inches of radius has an area of 2²π = 4π square inches, hence the probability of hitting that area is 4π/100π = 1/25 = 0.04

b) If the dart s within 2 inches of the rim, then it is not at distance 8 inches from the center (that is the complementary event). The probability for the dart to be at 8 inches of the center is 8²π/100π = 64/100 = 16/25 = 0.64, thus, the probability that the dart is at distance 2 or less from the rim is 1-0.64 = 0.36.

c) The first quadrant has an area exactly 4 times smaller than the area of the target (each quadrant has equal area), thus the probability for the dart to fall there is 1/4 = 0.25

d) If the dart is within 2 inches from the rim (which has probability 0.36 as we previously computed), then it will be equally likely for the dart to be in either of the 4 quadrants (the area that is within 2 inches from the rim forms a ring and it has equal area restricted on each quadrant). Therefore, the probability for the dice to be in the first qudrant and within 2 inches from the rim is 0.36*1/4 = 0.09.

6 0
4 years ago
Please help me with pythagoras.
lana [24]

Answer:

(q)

cb \:  =  \sqrt{4 {}^{2}  +  {3}^{2} }

cb = 5

q =  \sqrt{ {13}^{2}  -  {5}^{2} }

q = 12

(r)

cb = \sqrt{ {13}^{2} -  {5}^{2}  }

cb = 12

r =  \sqrt{12 {}^{2} - 5 {}^{2}  }

r =  \sqrt{119}

r = 10.91

(s)

ab =  \sqrt{20 {}^{2}  - 16 {}^{2} }

ab = 12

s =  \sqrt{12 {}^{2} + 8 {}^{2}  }

s = 4 \sqrt{13}

s = 14.42

Step-by-step explanation:

8 0
3 years ago
What is the value of the x in the equation? x+1/8=3/4​
creativ13 [48]

Answer:

x = 5/8

Step-by-step explanation:

Isolate the variable by subtracting 1/8 from both sides:

3/4 - 1/8 = 5/8

Now, the equation is:

x = 5/8

Therefore, the answer is 5/8

7 0
3 years ago
Read 2 more answers
Please help please giving out brainless
Readme [11.4K]

Answer:

C

Step-by-step explanation:

The length of arc AC (L) is calculated as

L = circumference of circle × fraction of circle

  = 2πr × \frac{150}{360}

  = 2π × 12 × \frac{5}{12}

 = 2π × 5

 = 10π ≈ 31.42 → C

4 0
3 years ago
Please help me. i will mark you as most brilliance and give you 5 stars! Please be specific when answering. thank you! :)
serg [7]

Step-by-step explanation:

5x+52=9x+20

subtract 5x on both sides

subtract 20 on both sides

32=4x

divide by 4 on both sides

x=8

plug in x to the equations for A, D, and C

A and D are 92°

C is 56°

B and C will add up to 90, so

90-56=34°

13y-6=34

add 6 on both sides

13y=40

divide by 13

y=40/13 (I don't wanna calculate that)

7 0
3 years ago
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