Answer:
![(g + f)(2) = \sqrt 3 - 3](https://tex.z-dn.net/?f=%28g%20%2B%20f%29%282%29%20%3D%20%5Csqrt%203%20-%203)
![(\frac{f}{g})(-1) = 0](https://tex.z-dn.net/?f=%28%5Cfrac%7Bf%7D%7Bg%7D%29%28-1%29%20%3D%200)
![(g + f)(-1) = \sqrt{15}](https://tex.z-dn.net/?f=%28g%20%2B%20f%29%28-1%29%20%3D%20%5Csqrt%7B15%7D)
![(g * f)(2) = -3\sqrt 3](https://tex.z-dn.net/?f=%28g%20%2A%20f%29%282%29%20%3D%20-3%5Csqrt%203)
Step-by-step explanation:
Given
![f(x) =1 - x^2](https://tex.z-dn.net/?f=f%28x%29%20%3D1%20-%20x%5E2)
![g(x) = \sqrt{11 - 4x](https://tex.z-dn.net/?f=g%28x%29%20%3D%20%5Csqrt%7B11%20-%204x)
See attachment
Solving (a): (g + f)(2)
This is calculated as:
![(g + f)(2) = g(2) + f(2)](https://tex.z-dn.net/?f=%28g%20%2B%20f%29%282%29%20%3D%20g%282%29%20%2B%20f%282%29)
Calculate g(2) and f(2)
![g(2) \to \sqrt{11 - 4 * 2} = \sqrt{3}](https://tex.z-dn.net/?f=g%282%29%20%5Cto%20%5Csqrt%7B11%20-%204%20%2A%202%7D%20%3D%20%5Csqrt%7B3%7D)
![f(2) = 1 - 2^2 = -3](https://tex.z-dn.net/?f=f%282%29%20%3D%201%20-%202%5E2%20%3D%20-3)
So:
![(g + f)(2) = g(2) + f(2)](https://tex.z-dn.net/?f=%28g%20%2B%20f%29%282%29%20%3D%20g%282%29%20%2B%20f%282%29)
![(g + f)(2) = \sqrt 3 - 3](https://tex.z-dn.net/?f=%28g%20%2B%20f%29%282%29%20%3D%20%5Csqrt%203%20-%203)
Solving (b): ![(\frac{f}{g})(-1)](https://tex.z-dn.net/?f=%28%5Cfrac%7Bf%7D%7Bg%7D%29%28-1%29)
This is calculated as:
![(\frac{f}{g})(-1) = \frac{f(-1)}{g(-1)}](https://tex.z-dn.net/?f=%28%5Cfrac%7Bf%7D%7Bg%7D%29%28-1%29%20%3D%20%5Cfrac%7Bf%28-1%29%7D%7Bg%28-1%29%7D)
Calculate f(-1) and g(-1)
![f(-1) = 1 - (-1)^2 = 0](https://tex.z-dn.net/?f=f%28-1%29%20%3D%201%20-%20%28-1%29%5E2%20%3D%200)
So:
![(\frac{f}{g})(-1) = \frac{f(-1)}{g(-1)}](https://tex.z-dn.net/?f=%28%5Cfrac%7Bf%7D%7Bg%7D%29%28-1%29%20%3D%20%5Cfrac%7Bf%28-1%29%7D%7Bg%28-1%29%7D)
![(\frac{f}{g})(-1) = \frac{0}{g(-1)}](https://tex.z-dn.net/?f=%28%5Cfrac%7Bf%7D%7Bg%7D%29%28-1%29%20%3D%20%5Cfrac%7B0%7D%7Bg%28-1%29%7D)
![(\frac{f}{g})(-1) = 0](https://tex.z-dn.net/?f=%28%5Cfrac%7Bf%7D%7Bg%7D%29%28-1%29%20%3D%200)
Solving (c): (g - f)(-1)
This is calculated as:
![(g + f)(-1) = g(-1) - f(-1)](https://tex.z-dn.net/?f=%28g%20%2B%20f%29%28-1%29%20%3D%20g%28-1%29%20-%20f%28-1%29)
Calculate g(-1) and f(-1)
![g(-1) = \sqrt{11 - 4 * -1} = \sqrt{15}](https://tex.z-dn.net/?f=g%28-1%29%20%3D%20%5Csqrt%7B11%20-%204%20%2A%20-1%7D%20%3D%20%5Csqrt%7B15%7D)
![f(-1) = 1 - (-1)^2 = 0](https://tex.z-dn.net/?f=f%28-1%29%20%3D%201%20-%20%28-1%29%5E2%20%3D%200)
So:
![(g + f)(-1) = g(-1) - f(-1)](https://tex.z-dn.net/?f=%28g%20%2B%20f%29%28-1%29%20%3D%20g%28-1%29%20-%20f%28-1%29)
![(g + f)(-1) = \sqrt{15} - 0](https://tex.z-dn.net/?f=%28g%20%2B%20f%29%28-1%29%20%3D%20%5Csqrt%7B15%7D%20-%200)
![(g + f)(-1) = \sqrt{15}](https://tex.z-dn.net/?f=%28g%20%2B%20f%29%28-1%29%20%3D%20%5Csqrt%7B15%7D)
Solving (d): (g * f)(2)
This is calculated as:
![(g * f)(2) = g(2) * f(2)](https://tex.z-dn.net/?f=%28g%20%2A%20f%29%282%29%20%3D%20g%282%29%20%2A%20f%282%29)
Calculate g(2) and f(2)
![g(2) \to \sqrt{11 - 4 * 2} = \sqrt{3}](https://tex.z-dn.net/?f=g%282%29%20%5Cto%20%5Csqrt%7B11%20-%204%20%2A%202%7D%20%3D%20%5Csqrt%7B3%7D)
![f(2) = 1 - 2^2 = -3](https://tex.z-dn.net/?f=f%282%29%20%3D%201%20-%202%5E2%20%3D%20-3)
So:
![(g * f)(2) = g(2) * f(2)](https://tex.z-dn.net/?f=%28g%20%2A%20f%29%282%29%20%3D%20g%282%29%20%2A%20f%282%29)
![(g * f)(2) = \sqrt 3 * -3](https://tex.z-dn.net/?f=%28g%20%2A%20f%29%282%29%20%3D%20%5Csqrt%203%20%2A%20-3)
![(g * f)(2) = -3\sqrt 3](https://tex.z-dn.net/?f=%28g%20%2A%20f%29%282%29%20%3D%20-3%5Csqrt%203)
Answer:
Step-by-step explanation:
![e=MC^2 \\ \frac{e}{C^2 } = M \\ \huge \red{ \boxed{M = \frac{e}{C^2 } }}](https://tex.z-dn.net/?f=e%3DMC%5E2%20%20%5C%5C%20%20%5Cfrac%7Be%7D%7BC%5E2%20%7D%20%20%3D%20M%20%5C%5C%20%5Chuge%20%5Cred%7B%20%5Cboxed%7BM%20%20%3D%20%5Cfrac%7Be%7D%7BC%5E2%20%7D%20%7D%7D)
Answer:
33124 passwords can be created using this format
Step-by-step explanation:
The password will have the following format:
L - L - D - D
There are 26 letters and 7 digits, so the number of possible outcomes in each position is:
26 - 26 - 7 - 7
How many passwords can be created using this format?
26*26*7*7 = 33124
33124 passwords can be created using this format
4^6(x^8)(y^-2) / 4^7(x^4)(y^2)
= 4^-1(x^4)(y^-4)
= x^4 / 4y^4
Therefore A is the correct answer. Obtained using exponent laws.
Laws used:
- Anything to the power of 0 is 1, which eliminates the base.
- When you divide like bases with exponents, you subtract the exponents to simplify.
Answer:
The percent is 75%
The amount is 45
The base is 60
Step-by-step explanation:
1) Set up the equation:
60x=45
2) Divide both sides by 60:
x=45/60
3) Simplify the fraction:
x=0.75
4) Write the decimal as a percent:
75%