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den301095 [7]
3 years ago
7

The question is about square root

Mathematics
2 answers:
Andrej [43]3 years ago
8 0
Well what you would do is find the square root of 4 and the square root of 10 and the square root of 4 is 2, and the sqaure root of 10 is 3.2. So I assume you can do the math from there
Ivahew [28]3 years ago
7 0
2/3.16=0.63 hope I help you
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What is the constant speed of freddy and sam?
dlinn [17]
The constant speed, or "average rate of change" is their slope.

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now, for Sam, let's use the last two points in the table then,

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6 0
3 years ago
Read 2 more answers
a trapezoid has a perimeter of 14ft what will the perimeter be if each side length id increased by a factor of 7
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5 0
2 years ago
Find the slope of the line that contains these points (-2,5) (-4,11)
nevsk [136]
The first thing you have to do is:

subtract: 11-5

Then subtract: -4-(-2)
 
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3 0
3 years ago
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A man starts walking north at 2 ft/s from a point P. Five minutes later a woman starts walking south at 3 ft/s from a point 500
Vedmedyk [2.9K]

Answer:

The rate at which both of them are moving apart is 4.9761 ft/sec.

Step-by-step explanation:

Given:

Rate at which the woman is walking,\frac{d(w)}{dt} = 3 ft/sec

Rate at which the man is walking,\frac{d(m)}{dt} = 2 ft/sec

Collective rate of both, \frac{d(m+w)}{dt} = 5 ft/sec

Woman starts walking after 5 mins so we have to consider total time traveled by man as (5+15) min  = 20 min

Now,

Distance traveled by man and woman are m and w ft respectively.

⇒ m=2\ ft/sec=2\times \frac{60}{min} \times 20\ min =2400\ ft

⇒ w=3\ ft/sec = 3\times \frac{60}{min} \times 15\ min =2700\  ft

As we see in the diagram (attachment) that it forms a right angled triangle and we have to calculate \frac{dh}{dt} .

Lets calculate h.

Applying Pythagoras formula.

⇒ h^2=(m+w)^2+500^2  

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Now differentiating the Pythagoras formula we can calculate the rate at which both of them are moving apart.

Differentiating with respect to time.

⇒ h^2=(m+w)^2+500^2

⇒ 2h\frac{d(h)}{dt}=2(m+w)\frac{d(m+w)}{dt}  + \frac{d(500)}{dt}

⇒ \frac{d(h)}{dt} =\frac{2(m+w)\frac{d(m+w)}{dt} }{2h}                         ...as \frac{d(500)}{dt}= 0

⇒ Plugging the values.

⇒ \frac{d(h)}{dt} =\frac{2(2400+2700)(5)}{2\times 5124.45}                       ...as \frac{d(m+w)}{dt} = 5 ft/sec

⇒ \frac{d(h)}{dt} =4.9761  ft/sec

So the rate from which man and woman moving apart is 4.9761 ft/sec.

3 0
3 years ago
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