1.) f(x)=7(b)^x-2
x=0→f(0)=7(b)^0-2=7(1)-2=7-2→f(0)=5→(x,f(x))=(0,5) Ok
2.) f(x)=-3(b)^x-5
x=0→f(0)=-3(b)^0-5=-3(1)-5=-3-5→f(0)=-8→(x,f(x))=(0,-8) No
3.) f(x)=5(b)^x-1
x=0→f(0)=5(b)^0-1=5(1)-1=5-1→f(0)=4→(x,f(x))=(0,4) No
4.) f(x)=-5(b)^x+10
x=0→f(0)=-5(b)^0+10=-5(1)+10=-5+10→f(0)=5→(x,f(x))=(0,5) Ok
5.) f(x)=2(b)^x+5
x=0→f(0)=2(b)^0+5=2(1)+5=2+5→f(0)=7→(x,f(x))=(0,7) No
Answers:
First option: f(x)=7(b)^x-2
Fourth option: f(x)=-5(b)^x+10
Answer:
i think its f
Step-by-step explanation:
Answer:

Step-by-step explanation:
Theoretically, there is not square root of neither 13 nor a negative number so the special symbol is used to represent the square root of a number.
Answer:
35.4 years
Step-by-step explanation:
The annual consumption (in billions of units) is described by the exponential function ...
f(t) = 45.5·1.026^t
The accumulated consumption is described by the integral ...

We want to find t such that the value of this integral is 2625, the estimated oil reserves.
2625 = 45.5/ln(1.026)·(1.026^t -1)
2625·ln(1.026)/45.5 +1 = 1.026^t ≈ 1.480832 +1 = 1.026^t
Taking natural logs, we have ...
ln(2.480832) = t·ln(1.026)
t ≈ ln(2.480832)/ln(1.026) ≈ 35.398
After about 35.4 years, the oil reserves will run out.