Answer:
1.4
Step-by-step explanation:
Sample size: 3
Lowest value: 2.89
Highest value: 4.29
Range: 1.4
formula for calculating the range:
Range = maximum(x_i) - minimum(x_i)
where x_i represents the set of values.
Answer:
k=24
Step-by-step explanation:
The tangent of the function f at x=a, can be found by differentiating f w.r.t. x and then replacing x with a.
f=-x^2+8x+20
Differentiating both sides:
f'=(-x^2+8x+20)'
By sum rule:
f'=(-x^2)'+(8x)'+(20)'
By constant multiple rule:
f'=-(x^2)'+8(x)'+(20)'
By constant rule:
f'=-(x^2)+8(x)'+0
By power rule:
f'=-2x+8
f' at x=a is -2a+8
This is the slope of any tangent line to the curve f.
The slope of g is 4 if you compare it to slope intercept form y=mx+b.
So we gave -2a+8=4.
Subtracr 8 on both sides: -2a=-4
Divide both sides by -2: a=2
The tangent line to the curve at x=2 is y=4x+k.
To tind y we must first know the y-coordinate of the point of tangency.
If x=2, then
f(2)=-(2)^2+8(2)+20=-4+16+20=12+20=32
So the point is (2,32).
g(x)=4x+k and we know g(2)=32.
This gives us:
32=4(2)+k
32=8+k
k=32-8
k=24
Answer:
a) ![v = \frac{[L]}{[T]} = LT^{-1}](https://tex.z-dn.net/?f=%20v%20%3D%20%5Cfrac%7B%5BL%5D%7D%7B%5BT%5D%7D%20%3D%20LT%5E%7B-1%7D)
b) ![a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}](https://tex.z-dn.net/?f=%20a%20%3D%20%5Cfrac%7B%5BL%7D%7BT%7D%5E%7B-1%7D%5D%7D%7B%7BT%7D%7D%3D%20L%20T%5E%7B-1%7D%20T%5E%7B-1%7D%3D%20L%20T%5E%7B-2%7D)
c) ![\int v dt = s(t) = [L]=L](https://tex.z-dn.net/?f=%20%5Cint%20v%20dt%20%3D%20s%28t%29%20%3D%20%5BL%5D%3DL)
d) ![\int a dt = v(t) = [L][T]^{-1}=LT^{-1}](https://tex.z-dn.net/?f=%20%5Cint%20a%20dt%20%3D%20v%28t%29%20%3D%20%5BL%5D%5BT%5D%5E%7B-1%7D%3DLT%5E%7B-1%7D)
e) ![\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bda%7D%7Bdt%7D%3D%20%5Cfrac%7B%5BL%5D%5BT%5D%5E%7B-2%7D%7D%7BT%7D%20%3D%20%5BL%5D%5BT%5D%5E%7B-2%7D%20%5BT%5D%5E%7B-1%7D%20%3D%20LT%5E%7B-3%7D)
Step-by-step explanation:
Let define some notation:
[L]= represent longitude , [T] =represent time
And we have defined:
s(t) a position function
Part a
If we do the dimensional analysis for v we got:
Part b
For the acceleration we can use the result obtained from part a and we got:
Part c
From definition if we do the integral of the velocity respect to t we got the position:
And the dimensional analysis for the position is:
Part d
The integral for the acceleration respect to the time is the velocity:
And the dimensional analysis for the position is:
Part e
If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:
Answer:
It is located in the 3rd quadrant since x = -3 and y = -4.
The answer is 12 not to be fr
