Pythagorean Theorem<h2>
Verbally:</h2>
Let's say a and b are the legs, and c is the hypotenuse. Then, algebraically, the theorem is,
Answer:
46.9 for the first one
Step-by-step explanation:
So sorry if i'm wrong <3
Answer:
Hope this helps
Step-by-step explanation:
1/4x = -22 + 18
1/4x = -4
4(1/4x) = 4(-4)
x = -16
Answer:
![\dfrac{-1}{x(x+h)}, h\ne 0](https://tex.z-dn.net/?f=%5Cdfrac%7B-1%7D%7Bx%28x%2Bh%29%7D%2C%20h%5Cne%200)
Step-by-step explanation:
If
, then
. It follows that
![\begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \cdot [g(x+h) - g(x)] \\&= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%20%5C%5C%5Cfrac%7Bg%28x%2Bh%29-g%28x%29%7D%7Bh%7D%20%26%3D%20%5Cfrac%7B1%7D%7Bh%7D%20%5Ccdot%20%5Bg%28x%2Bh%29%20-%20g%28x%29%5D%20%5C%5C%26%3D%20%5Cfrac%7B1%7D%7Bh%7D%20%5Cleft%28%20%5Cfrac%7B1%7D%7Bx%2Bh%7D%20-%20%5Cfrac%7B1%7D%7Bx%7D%20%5Cright%29%5Cend%7Baligned%7D)
Technically we are done, but some more simplification can be made. We can get a common denominator between 1/(x+h) and 1/x.
![\begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\\&=\frac{1}{h} \left(\frac{x}{x(x+h)} - \frac{x+h}{x(x+h)} \right) \\ &=\frac{1}{h} \left(\frac{x-(x+h)}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{x-x-h}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{-h}{x(x+h)}\right) \end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%20%5C%5C%5Cfrac%7Bg%28x%2Bh%29-g%28x%29%7D%7Bh%7D%20%26%3D%20%5Cfrac%7B1%7D%7Bh%7D%20%5Cleft%28%20%5Cfrac%7B1%7D%7Bx%2Bh%7D%20-%20%5Cfrac%7B1%7D%7Bx%7D%20%5Cright%29%5C%5C%26%3D%5Cfrac%7B1%7D%7Bh%7D%20%5Cleft%28%5Cfrac%7Bx%7D%7Bx%28x%2Bh%29%7D%20-%20%5Cfrac%7Bx%2Bh%7D%7Bx%28x%2Bh%29%7D%20%5Cright%29%20%5C%5C%20%26%3D%5Cfrac%7B1%7D%7Bh%7D%20%5Cleft%28%5Cfrac%7Bx-%28x%2Bh%29%7D%7Bx%28x%2Bh%29%7D%5Cright%29%20%5C%5C%20%26%3D%5Cfrac%7B1%7D%7Bh%7D%20%5Cleft%28%5Cfrac%7Bx-x-h%7D%7Bx%28x%2Bh%29%7D%5Cright%29%20%5C%5C%20%26%3D%5Cfrac%7B1%7D%7Bh%7D%20%5Cleft%28%5Cfrac%7B-h%7D%7Bx%28x%2Bh%29%7D%5Cright%29%20%5Cend%7Baligned%7D)
Now we can cancel the h in the numerator and denominator under the assumption that h is not 0.
![= \dfrac{-1}{x(x+h)}, h\ne 0](https://tex.z-dn.net/?f=%3D%20%5Cdfrac%7B-1%7D%7Bx%28x%2Bh%29%7D%2C%20h%5Cne%200)