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kicyunya [14]
8 months ago
15

Which equation of a straight line is parallel to the line 4y= 3x +5? A) y= 3x +5. B) 4y= 3x-1. C) 3y= 4x +5. D) -4y= 3x+5. E) 4y

= x + 5
Mathematics
1 answer:
kiruha [24]8 months ago
5 0

Answer:

The answer is 3y= 4x +5

Step-by-step explanation:

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2 years ago
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Math help.... Please....
garri49 [273]
First, you want to solve for the equation for this problem, would be: 
0.05N + 0.10D = 20.50 
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Since N = 164, and D = 123. It would add up to 287 coins. 164 + 123 = 287. 
Now that you have the number for those two variables, solve the equation for when N = 164, and D = 123. 
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</span><span>8.2 + 12.3 = 20.50 
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<em>I hope this helps! </em>

<em>~ Notorious Sovereign</em>
6 0
3 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!

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2 years ago
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