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shtirl [24]
3 years ago
11

Can someone please help me solve this. So confusing

Mathematics
1 answer:
borishaifa [10]3 years ago
4 0

Step-by-step explanation:

\frac{a}{ {r}^{3} }  + h = d

\frac{a}{ {r}^{3} }  = d - h

a = (d - h) \times  {r}^{3}

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Help I don’t understand the concept
kvasek [131]

Answer:

C. No, because one x-value corresponds to two different y-values

Step-by-step explanation:

An x-value cannot have two y-values. Hope this helps!

6 0
3 years ago
Read 2 more answers
A chalk board’s dimensions have an
Wewaii [24]

Step-by-step explanation:

5:3=35

5=35×3

5=105

5÷105=0.05

7 0
2 years ago
Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'
vodka [1.7K]

Answer:

a) v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9

a(t) = x''(t) = v'(t) =6t-12

b)  0

c) a(t) = x''(t) = v'(t) =6t-12

When the acceleration is 0 we have:

6t-12=0, t =2

And if we replace t=2 in the velocity function we got:

v(t) = 3(2)^2 -12(2) +9=-3

Step-by-step explanation:

For this case we have defined the following function for the position of the particle:

x(t) = t^3 -6t^2 +9t -5 , 0\leq t\leq 10

Part a

From definition we know that the velocity is the first derivate of the position respect to time and the accelerations is the second derivate of the position respect the time so we have this:

v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9

a(t) = x''(t) = v'(t) =6t-12

Part b

For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:

v(t) = 3t^2 -12t +9

We can factorize this function as v(t)= 3 (t^2- 4t +3)=3(t-3)(t-1)

So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is 0\leq t\leq 10 we need to analyze the following intervals:

0< t

For this case if we select two values let's say 0.25 and 0.5 we see that

v(0.25) =6.1875, v(0.5)=3.75

And we see that for a=0.5 >0.25=b we have that f(b)>f(a) so then the function is decreasing on this case.  

1

We have a minimum at t=2 since at this value w ehave the vertex of the parabola :

v_x =-\frac{b}{2a}= -\frac{-12}{2*3}= -2

And at t=-2 v(2) = -3 that represent the minimum for this function, we see that if we select two values let's say 1.5 and 1.75

v(1.75) =-2.8125< -2.25= v(1.5) so then the function sis decreasing on the interval 1<t<2

2

We see that the function would be increasing.

3

For this interval we will see that for any two points a,b with a>b we have f(a)>f(b) for example let's say a=3 and b =4

f(a=3) =0 , f(b=4) =9 , f(b)>f(a)

The particle is moving to the right then the velocity is positive so then the answer for this case is: 0

Part c

a(t) = x''(t) = v'(t) =6t-12

When the acceleration is 0 we have:

6t-12=0, t =2

And if we replace t=2 in the velocity function we got:

v(t) = 3(2)^2 -12(2) +9=-3

5 0
3 years ago
If f(x)= x^-2, and g(x)=3x-1 whats the domain of f(g(x))?
andrey2020 [161]
4 and 7 and 8 and 9 and10
5 0
3 years ago
20 POINTS!!! I GIVE BRAINLIEST PLEASE HELP!!!!A circle with a circumference of approximately 15.71 inches drawn inside a square,
Oksi-84 [34.3K]

answer:

5.002 inches

step-by-step explanation:

  • find the diameter of the circle
  • the diameter is the side length of the square since the sides of it are touching the circle

c = 2πr

  • c stands for circumference
  • π stands for pi (or 3.14)
  • r stands for radius

- plug in the values given to find the diameter

c = 2πr

15.71 = 2(3.14)r

15.71 = 6.28r

r = 2.501 (rounded or estimated)

  • now that we have the radius, we can find the diameter by multiplying by 2, since diameter is two times bigger than the radius

diameter = 5.002 (estimated)

  • now, that will be the side length because the diameter = the side length
3 0
2 years ago
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