Answer:
367441
Step-by-step explanation:
When the velocity goes from 40km/h to 20 km/h, the kinetic energy decreases by a factor of 4.
<h3>
What happens to the kinetic energy?</h3>
We know that the kinetic energy depends of the square of the velocity. Thus, if we decrease the velocity from 40km/h to 20km/h, then the kinetic energy decreases.
Remember that the kinetic energy is:
K = (m/2)*v²
Where m is the mass.
The initial kinetic energy is:
K = (m/2)*(40km/h)²
The final kinetic energy is:
K' = (m/2)*(20km/h)²
The quotient gives:
K/K' = [ (m/2)*(40km/h)²]/[ (m/2)*(20km/h)²]
K/K' = (40km/h)²/(20km/h)² = 4
So the kinetic energy decreases by a factor of 4.
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The following are true:
<span>The graph has a relative minimum
The graph of the quadratic function has a vertex at (0,5)
The graph opens up
The axis of symmetry is x=0
It has a minimum point; there is a smallest value of y for this function. The vertex will be that minimum point, (0, 5). Since that is the vertex, that means x=0 is the axis of symmetry. It opens up because all of the other points are above the vertex.
It does not have any x-intercepts, since the lowest y-value is 5 (the vertex). The y-intercept would be at (0, 5), not (5, 0).</span>
The values of x in the equation 1/3 * |x - 3| + 4 = 10 are x = 21 and x = -15
<h3>How to solve for x in the equation?</h3>
The equation statement is given as:
one third times the absolute value of the quantity x minus 3 end quantity plus 4 equals 10
Rewrite the equation as follows:
1/3 * |x - 3| + 4 = 10
Subtract 4 from both sides of the equation
1/3 * |x - 3| = 6
Multiply both sides of the equation by 3
|x - 3| = 18
Expand the equation
x - 3 = 18 and 3 - x = 18
Solve for x in both equations
x = 21 and x = -15
Hence, the values of x in the equation 1/3 * |x - 3| + 4 = 10 are x = 21 and x = -15
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Answer:
f(2) = -6
f inverse of 1/2 is 17/2
Step-by-step explanation:
We are given a function ![\displaystyle \large{f(x) = \frac{6}{2x-5}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clarge%7Bf%28x%29%20%3D%20%5Cfrac%7B6%7D%7B2x-5%7D%7D)
To solve for part (a), we have to substitute x = 2 in f(x).
![\displaystyle \large{f(2)=\frac{6}{2(2)-5} \\\displaystyle \large{f(2)=\frac{6}{4-5}=\frac{6}{-1}=-6](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clarge%7Bf%282%29%3D%5Cfrac%7B6%7D%7B2%282%29-5%7D%20%5C%5C%5Cdisplaystyle%20%5Clarge%7Bf%282%29%3D%5Cfrac%7B6%7D%7B4-5%7D%3D%5Cfrac%7B6%7D%7B-1%7D%3D-6)
Therefore, f(2) is -6.
Now we solve for part (b), we see the notation which is a minor different similar to f(x). We see that there is exponent of -1 between f and 1/2. The part (b) indicates that the function is an inverse of f(x).
![\displaystyle \large{y=f^{-1}(x) \longrightarrow x=f(y)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clarge%7By%3Df%5E%7B-1%7D%28x%29%20%5Clongrightarrow%20x%3Df%28y%29%7D)
To solve for part (b), first we solve for x in the function. Let f(x) = y.
![\displaystyle \large{y=\frac{6}{2x-5}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clarge%7By%3D%5Cfrac%7B6%7D%7B2x-5%7D%7D)
Multiply both sides by 2x-5.
![\displaystyle \large{y(2x-5)=\frac{6}{2x-5}(2x-5)}\\\displaystyle \large{y(2x-5)=6}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clarge%7By%282x-5%29%3D%5Cfrac%7B6%7D%7B2x-5%7D%282x-5%29%7D%5C%5C%5Cdisplaystyle%20%5Clarge%7By%282x-5%29%3D6%7D)
Divide both sides by y-term.
![\displaystyle \large{\frac{y(2x-5)}{y}=\frac{6}{y}}\\\displaystyle \large{2x-5=\frac{6}{y}}\\\displaystyle \large{2x=\frac{6}{y}+5}\\\displaystyle \large{x=\frac{6}{2y}+\frac{5}{2}}\\\displaystyle \large{x=\frac{3}{y}+\frac{5}{2}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clarge%7B%5Cfrac%7By%282x-5%29%7D%7By%7D%3D%5Cfrac%7B6%7D%7By%7D%7D%5C%5C%5Cdisplaystyle%20%5Clarge%7B2x-5%3D%5Cfrac%7B6%7D%7By%7D%7D%5C%5C%5Cdisplaystyle%20%5Clarge%7B2x%3D%5Cfrac%7B6%7D%7By%7D%2B5%7D%5C%5C%5Cdisplaystyle%20%5Clarge%7Bx%3D%5Cfrac%7B6%7D%7B2y%7D%2B%5Cfrac%7B5%7D%7B2%7D%7D%5C%5C%5Cdisplaystyle%20%5Clarge%7Bx%3D%5Cfrac%7B3%7D%7By%7D%2B%5Cfrac%7B5%7D%7B2%7D%7D)
Then swap x and y which we receive ![\displaystyle \large{y=\frac{3}{x}+\frac{5}{2}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clarge%7By%3D%5Cfrac%7B3%7D%7Bx%7D%2B%5Cfrac%7B5%7D%7B2%7D%7D)
Therefore, ![\displaystyle \large{f(x)=\frac{6}{2x-5} \longrightarrow f^{-1}(x)=\frac{3}{x}+\frac{5}{2}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clarge%7Bf%28x%29%3D%5Cfrac%7B6%7D%7B2x-5%7D%20%5Clongrightarrow%20f%5E%7B-1%7D%28x%29%3D%5Cfrac%7B3%7D%7Bx%7D%2B%5Cfrac%7B5%7D%7B2%7D%7D)
Thus, ![\displaystyle \large{f^{-1}(\frac{1}{2})=\frac{3}{\frac{1}{2}}+\frac{5}{2}}\\\displaystyle \large{f^{-1}(\frac{1}{2})=6+\frac{5}{2}}\\\displaystyle \large{f^{-1}(\frac{1}{2})=\frac{17}{2}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clarge%7Bf%5E%7B-1%7D%28%5Cfrac%7B1%7D%7B2%7D%29%3D%5Cfrac%7B3%7D%7B%5Cfrac%7B1%7D%7B2%7D%7D%2B%5Cfrac%7B5%7D%7B2%7D%7D%5C%5C%5Cdisplaystyle%20%5Clarge%7Bf%5E%7B-1%7D%28%5Cfrac%7B1%7D%7B2%7D%29%3D6%2B%5Cfrac%7B5%7D%7B2%7D%7D%5C%5C%5Cdisplaystyle%20%5Clarge%7Bf%5E%7B-1%7D%28%5Cfrac%7B1%7D%7B2%7D%29%3D%5Cfrac%7B17%7D%7B2%7D%7D)
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