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:
3/5
Step-by-step explanation:
There are 20 balls, numbered 1 through 20.
Of these, 10 are even, 4 are less than five, and 2 are both even and less than 5.
So the probability is:
P(even or <5) = P(even) + P(<5) − P(even and <5)
P(even or <5) = 10/20 + 4/20 − 2/20
P(even or <5) = 12/20
P(even or <5) = 3/5
Answer:
The resulting three-dimensional object will be a cone
Step-by-step explanation:
Plot the figure to better understand the problem'
we have
a triangular cross-section with coordinates at (1, 1), (1, 4), and (3, 1)
The draw in the attached figure
we know that
When rotating a a right triangle about a leg, the three dimensional figure formed is a cone
The radius of the cone will be equal to the base of right triangle and the height of the cone will be equal to the height of the triangle
so
The radius of the cone will be equal to 2 units and the height of the cone will be 3 units
see the attached figure to better understand the problem