These operation are called reverse operation......... I hope this helps
We grapht hem
we see that the base is from (-2,2) and (6,2), 8 units long
the height will be from y=2 to y=8 or 6 units
so then, the area=1/2 times base times height=1/2 times 8 times 6=24 square units
The inverse is the symmetrical function relatively to the function y=x. So, it basically means, that to find the inverse, you just need:
1. Switch x and y (or f(x)): x = 5H(x) + 2
2. Solve for y again (or for f(x)):
5H(x) = x -2
H(x) = (x-2)/5
This is the inverse.
Answer:
Option D. ![H(t) = 6.5+21t-16t^2](https://tex.z-dn.net/?f=H%28t%29%20%3D%206.5%2B21t-16t%5E2)
Step-by-step explanation:
we know that
If air resistance is ignored, the height h (in feet) of the ball t seconds after it is served is given by
![H(t) = - \frac{1}{2}g t^2 + V_0t + H_0](https://tex.z-dn.net/?f=H%28t%29%20%3D%20-%20%5Cfrac%7B1%7D%7B2%7Dg%20t%5E2%20%2B%20V_0t%20%2B%20H_0)
where
g is the acceleration due to gravity which on earth is approximately equal to 32 feet / sec^2
V_0 is the initial velocity (when t = 0 )
H_0 is the initial height (when t = 0)
In this problem we have
![V_0=21\ ft/sec\\H_0=6.5\ ft](https://tex.z-dn.net/?f=V_0%3D21%5C%20ft%2Fsec%5C%5CH_0%3D6.5%5C%20ft)
substitute the given values
![H(t) = - \frac{1}{2}(32) t^2 + (21)t + 6.5](https://tex.z-dn.net/?f=H%28t%29%20%3D%20-%20%5Cfrac%7B1%7D%7B2%7D%2832%29%20t%5E2%20%2B%20%2821%29t%20%2B%206.5)
![H(t) = -16t^2 + 21t + 6.5](https://tex.z-dn.net/?f=H%28t%29%20%3D%20-16t%5E2%20%2B%2021t%20%2B%206.5)
Rewrite
![H(t) = 6.5+21t-16t^2](https://tex.z-dn.net/?f=H%28t%29%20%3D%206.5%2B21t-16t%5E2)
Answer:
Step-by-step explanation:
We have the equations
4x + 3y = 18 where x = the side of the square and y = the side of the triangle
For the areas:
A = x^2 + √3y/2* y/2
A = x^2 + √3y^2/4
From the first equation x = (18 - 3y)/4
So substituting in the area equation:
A = [ (18 - 3y)/4]^2 + √3y^2/4
A = (18 - 3y)^2 / 16 + √3y^2/4
Now for maximum / minimum area the derivative = 0 so we have
A' = 1/16 * 2(18 - 3y) * -3 + 1/4 * 2√3 y = 0
-3/8 (18 - 3y) + √3 y /2 = 0
-27/4 + 9y/8 + √3y /2 = 0
-54 + 9y + 4√3y = 0
y = 54 / 15.93
= 3.39 metres
So x = (18-3(3.39) / 4 = 1.96.
This is a minimum value for x.
So the total length of wire the square for minimum total area is 4 * 1.96
= 7.84 m
There is no maximum area as the equation for the total area is a quadratic with a positive leading coefficient.