1. Consider y=f(x)
2. let a and b be 2 positive numbers
3. then the graph of y=f(x)-a is the graph of y=f(x) shifted a units down
4. the graph of y=f(x+b) is y=f(x) shifted b units left
y=f(x-b) is y=f(x) shifted b units right
5. y=f(x+b)-a is the graph of y=f(x) shifted a units down and b units left
6. So
shifted 4 units down and 5 units left is the graph of
7. To check: consider
at x=3. We have the point (3, 27)
Shift this point 4 units down and 5 units left: (3-5, 27-4)=(-2, 23)
Consider
for x=-2
According to the formula above
7100=1300 (1+0.075)^t
Solve for t
Divide both sides by 1300 to get
7100/1300=(1+0.075)^t
Take the log for both sides
Log (7100/1300)=t×log (1.075)
Divide both sides by log (1.075) to get
T=log(7,100÷1,300)÷log(1.075)
T=23.48 years round your answer to get 23 years
Hope it helps!
Answer: f(y) = (Cot(y) + 2/3)*(4/3)
Step-by-step explanation:
We have the function:
y = Arccot( 3*x/4 - 2/3)
And we want to find the inverse, then we must isolate x in the right side.
Applying cot( ) to both sides, we get
Cot( y) = Cot(Arccot( 3*x/4 - 2/3)) = 3*x/4 - 2/3
Cot(y) = 3*x/4 - 2/3
Cot(y) + 2/3 = 3*x/4
(Cot(y) + 2/3)*(4/3) = x
Then the inverse function is:
f(y) = (Cot(y) + 2/3)*(4/3)
B is a perfect cube Bc all add up to five