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kozerog [31]
4 years ago
13

Please solve 39 and 41

Mathematics
1 answer:
s344n2d4d5 [400]4 years ago
5 0
 dexter dexter dexter dexter dexter dexter dexter dexter<span> dexter</span>
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Х у
erica [24]

Answer:

A. Linear

Step-by-step explanation:

As the x value goes up by 1, the y value goes up by 3 each time

8 0
2 years ago
Leah practices dance for 2 hours each week. Her younger sister Sonia practices dance for 0.8 the amount of time Leah practices.
galben [10]
Sonia practices for 96 mins (1 hour 36mins ) 0.8 = 80%
3 0
3 years ago
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If a gas tank of a large truck holds 64 gallons of gas if the tank is 5/8 full how many gallons of gas is in truck
den301095 [7]
Find 5/8 of full
find 5/8 of 64
5/8 times 64/1=320/8=160/4=80/2=40
answer is 40 gallons in the trick right now
4 0
3 years ago
Calculate the number of two-person committees that can be formed from a group of 10 people
ioda

Bikash borrowed Rs150,000 from Anish at the rare 21% per year at the end of 9 month. How compound interest should he pay compound half yearly.

4 0
2 years ago
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
3 years ago
Read 2 more answers
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