Answer:
The value of x that maximizes the volume enclosed by this box is 0.46 inches
The maximum volume is 3.02 cubic inches
Step-by-step explanation:
see the attached figure to better understand the problem
we know that
The volume of the open-topped box is equal to

where

substitute

Convert to expanded form

using a graphing tool
Graph the cubic equation
Remember that
The domain for x is the interval -----> (0,1)
Because
If x>1
then
the width is negative (W=2-2x)
so
The maximum is the point (0.46,3.02)
see the attached figure
therefore
The value of x that maximizes the volume enclosed by this box is 0.46 inches
The maximum volume is 3.02 cubic inches
Option b. 0.496
- Step-by-step explanation:
we know that
f(x)=0.01(2^{x})
The average rate of change is equal to
\frac{f(b)-f(a)}{b-a}
where
a=3
b=8
f(a)=f(3)=0.01(2^{3})=0.08
f(b)=f(8)=0.01(2^{8})=2.56
Substitute
\frac{2.56-0.08}{8-3}
\frac{2.48}{5}
0.496
It’s d because it is a negative slope and the b intercept is negative so it never is in the first quadrant
5 consecutive integers : x, x + 1, x + 2, x + 3, x + 4
x + x + 1 + x + 2 + x + 3 + x + 4 = 505
5x + 10 = 505
5x = 505 - 10
5x = 495
x = 495/5
x = 99
x + 1 = 99 + 1 = 100
x + 2 = 99 + 2 = 101 <=== 3rd number is 101
x + 3 = 99 + 3 = 102
x + 4 = 99 + 4 = 103
I think it’d be 2 but I’m not really sure