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emmainna [20.7K]
3 years ago
15

How do you do these two questions?

Mathematics
1 answer:
Mice21 [21]3 years ago
4 0

Answer:

a) d²y/dx² = ½ x + y − ½

b) Relative minimum

Step-by-step explanation:

a) Take the derivative with respect to x.

dy/dx = ½ x + y − 1

d²y/dx² = ½ + dy/dx

d²y/dx² = ½ + (½ x + y − 1)

d²y/dx² = ½ x + y − ½

b) At (0, 1), the first and second derivatives are:

dy/dx = ½ (0) + (1) − 1

dy/dx = 0

d²y/dx² = ½ (0) + (1) − ½

d²y/dx² = ½

The first derivative is 0, and the second derivative is positive (concave up).  Therefore, the point is a relative minimum.

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While passing a slower car on the highway, i accelerate uniformly from 12 m/s to 24 m/s in a time of 10.0s. How far do you trave
Taya2010 [7]
You say that there is uniform acceleration so:

vf-vi=at  (final velocity minus initial velocity is equal to acceleration times time)

We know vf, vi, and t so we can solve for acceleration:

24-12=a10

12=10a

a=1.2

That is the acceleration, we will need to integrate with respect to time twice...

v=⌠a dt

v=at+vi  , we know a=1.2m/s^2 and vi=12m/s

v=1.2t+12, 

x=⌠1.2t+12 dt

x=1.2t^2/2+12t+xo, we can just let xo=0 for this problem...

x(t)=0.6t^2+12t

Now we know that this acceleration lasts for 10 seconds so the distance traveled in that time is:

x(10)=0.6(10^2)+12(10)

x(10)=60+120

x(10)=180 meters
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Step-by-step explanation:

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Could someone help me out
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It is the second one
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