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zysi [14]
2 years ago
15

I know i chose an answer but pls answer

Mathematics
1 answer:
const2013 [10]2 years ago
7 0
It’s b! 803.84 feet! Have a great day
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BRAINLIEST!!!<br> 18. Point p is chosen at random on EH. Find the probability that p is on FG.
shepuryov [24]

Answer:

\frac{2}{5}

Step-by-step explanation:

The total length of this line is 15 units. FG is 6 units long.

This means that the probability of p being on FG would be \frac{6}{15} which can be simplified to \frac{2}{5}

5 0
3 years ago
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⅔ (5/16)=equal too ??
Zepler [3.9K]

This is the answer

.... <em>please</em><em> mark</em><em> brainliest</em>

5 0
3 years ago
Can someone help?
nydimaria [60]

Answer:

run = 1081.6 ft

Step-by-step explanation:

sin = opp/hyp

sin18 = 3500 / run

run = 3500 / sin18

run = 1081.6 ft

4 0
2 years ago
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Solve the equation 3x - 2(x + 1) = 2x - 7
natta225 [31]

Answer:

X=5

Step-by-step explanation:

Collect like terms then move the terms. Then collect liked terms. Calculate and Chang signs

3 0
2 years ago
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A baseball diamond is a square with side 90 ft. A batter hits the ball and runs toward first base with a speed of 31 ft/s. (a) A
julsineya [31]

Answer:

a) -13.9 ft/s

b) 13.9 ft/s

Step-by-step explanation:

a) The rate of his distance from the second base when he is halfway to first base can be found by differentiating the following Pythagorean theorem equation respect t:

D^{2} = (90 - x)^{2} + 90^{2}   (1)

\frac{d(D^{2})}{dt} = \frac{d(90 - x)^{2} + 90^{2})}{dt}

2D\frac{d(D)}{dt} = \frac{d((90 - x)^{2})}{dt}  

D\frac{d(D)}{dt} = -(90 - x) \frac{dx}{dt}   (2)

Since:

D = \sqrt{(90 -x)^{2} + 90^{2}}

When x = 45 (the batter is halfway to first base), D is:

D = \sqrt{(90 - 45)^{2} + 90^{2}} = 100. 62

Now, by introducing D = 100.62, x = 45 and dx/dt = 31 into equation (2) we have:

100.62 \frac{d(D)}{dt} = -(90 - 45)*31          

\frac{d(D)}{dt} = -\frac{(90 - 45)*31}{100.62} = -13.9 ft/s

Hence, the rate of his distance from second base decreasing when he is halfway to first base is -13.9 ft/s.

b) The rate of his distance from third base increasing at the same moment is given by differentiating the folowing Pythagorean theorem equation respect t:

D^{2} = 90^{2} + x^{2}  

\frac{d(D^{2})}{dt} = \frac{d(90^{2} + x^{2})}{dt}

D\frac{dD}{dt} = x\frac{dx}{dt}   (3)

We have that D is:

D = \sqrt{x^{2} + 90^{2}} = \sqrt{(45)^{2} + 90^{2}} = 100.63

By entering x = 45, dx/dt = 31 and D = 100.63 into equation (3) we have:

\frac{dD}{dt} = \frac{45*31}{100.63} = 13.9 ft/s

Therefore, the rate of the batter when he is from third base increasing at the same moment is 13.9 ft/s.

I hope it helps you!

4 0
2 years ago
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