Answer:
#UseDistributiveProperty
Step-by-step explanation:
y = 20! always remember that it's always (x,y)!
Answer:
3 2/5 cups remaining
Step-by-step explanation:
To subtract fractions, they need a common denominator. I am using 5.
7 is equal to 35/5. Make 3-3/5 into an improper fraction of 18/5.
Now subtract: 35 - 18 = 17. There is 17/5 remaining.
Make it into a mixed number and you will get 3 2/5 remaining.
Answer:
Option b
Step-by-step explanation:
To solve this problem we must test if the function is even.
If f(-x) = f(x) then the function is even and is symmetric with respect to the y-axis.
If f(-x) = -f(x) then the function is odd and has symmetry with respect to the origin.
We have the function:
![f(x) = \frac{e^x + e^{-x}}{2}](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7Be%5Ex%20%2B%20e%5E%7B-x%7D%7D%7B2%7D)
We make:
![f(-x) = \frac{e^{-x} + e^{-(-x)}}{2}](https://tex.z-dn.net/?f=f%28-x%29%20%3D%20%5Cfrac%7Be%5E%7B-x%7D%20%2B%20e%5E%7B-%28-x%29%7D%7D%7B2%7D)
Rewriting the function we have
![f(-x) = \frac{e^{-x} + e^{x}}{2} = \frac{e^{x} + e^{-x}}{2}\\\\f(-x) = f(x)](https://tex.z-dn.net/?f=f%28-x%29%20%3D%20%5Cfrac%7Be%5E%7B-x%7D%20%2B%20e%5E%7Bx%7D%7D%7B2%7D%20%3D%20%5Cfrac%7Be%5E%7Bx%7D%20%2B%20e%5E%7B-x%7D%7D%7B2%7D%5C%5C%5C%5Cf%28-x%29%20%3D%20f%28x%29)
Finally, the function has symmetry with respect to the y axis.
Case a)
f(x)=[x-1]/[x+5]
step 1
f(x)=y
y=[x-1]/[x+5]
step 2
exchange x for y and y for x
y=[x-1]/[x+5]------> x=[y-1]/[y+5]----> x*[y+5]=[y-1]----> xy+5x=y-1
step 3
clear the variable y
xy+5x=y-1-----> y-xy=5x+1----> y*[1-x]=[5x+1]----> y=[5x+1]/[1-x]
step 4
f(x)-1= [5x+1]/[1-x]
the function and the inverse function are not the same
case b)
g(x)=[x-2]/[x-1]
step 1
g(x)=y
y=[x-2]/[x-1]
step 2
exchange x for y and y for x
y=[x-2]/[x-1]------> x=[y-2]/[y-1]----> x*[y-1]=[y-2]----> xy-x=y-2
step 3
clear the variable y
xy-x=y-2-----> xy-y=-2+x----> y*[x-1]=[x-2]----> y=[x-2]/[x-1]
step 4
g(x)-1= [x-2]/[x-1]
the function and the inverse function are the same
case c)
h(x)=[x+3]/[x-2]
step 1
h(x)=y
y=[x+3]/[x-2]
step 2
exchange x for y and y for x
y=[x+3]/[x-2]------> x=[y+3]/[y-2]----> x*[y-2]=[y+3]----> xy-2x=y+3
step 3
clear the variable y
xy-2x=y+3-----> xy-y=3+2x----> y*[x-1]=[2x+3]----> y=[2x+3]/[x-1]
step 4
h(x)-1= [2x+3]/[x-1]
the function and the inverse function are not the same
case d)
k(x)=[x+1]/[x-1]
step 1
k(x)=y
y=[x+1]/[x-1]
step 2
exchange x for y and y for x
y=[x+1]/[x-1]------> x=[y+1]/[y-1]---> x*[y-1]=[y+1]----> xy-x=y+1
step 3
clear the variable y
xy-x=y+1-----> xy-y=x+1----> y*[x-1]=[x+1]----> y=[x+1]/[x-1]
step 4
k(x)-1= [x+1]/[x-1]
the function and the inverse function are the same