The function is written as:
f(x) = log(-20x + 12√x)
To find the maximum value, differentiate the equation in terms of x, then equate it to zero. The solution is as follows.
The formula for differentiation would be:
d(log u)/dx = du/u ln(10)
Thus,
d/dx = (-20 + 6/√x)/(-20x + 12√x)(ln 10) = 0
-20 + 6/√x = 0
6/√x = 20
x = (6/20)² = 9/100
Thus,
f(x) = log(-20(9/100)+ 12√(9/100)) = 0.2553
<em>The maximum value of the function is 0.2553.</em>
Answer:
The time-mean speed of the minivans is of 105.8 seconds.
Step-by-step explanation:
Mean of a data-set:
The mean of a data-set is the sum of all values in the data-set divided by the number of values.
Five minivans, times of: 98.0, 108.0, 113.0, 108.0, 102.0, in seconds.
Thus, the mean is:

The time-mean speed of the minivans is of 105.8 seconds.
Answer:
6 cm
Step-by-step explanation:
If the linear scale factor of two solids is k, then the volume scale factor is k^3.
The volume scale factor is 128/54 = 64/27 = (4/3)^3.
The linear scale factor is 4/3.
4.5 cm * 4/3 = 6 cm
Answer: The height of the larger container is 6 cm.
Answer:
P(N)=0.3
Step-by-step explanation:
Given: P(M)= 0.46, P(M and N)=0.138
Using P(M) ×P(N)= P(M and N)
⇒0.46×P(N)= 0.138
⇒P(N)= 
⇒P(N)=0.3
Answer:
3rd option: B(C)= 1.79C +86.03
Step-by-step explanation:
Total bill
= cost of cans(number of cans) +cost of other groceries
Let the cost of other groceries be G, and the cost of cans be X.
Given that number of cans= C,
Total bill= XC +G
If 2 cans were purchased,
2X+G= 89.61 -----(1)
If 5 cans were purchased,
5X +G= 94.98 -----(2)
(2) -(1):
(5X +G) -(2X +G)= 94.98 -89.61
5X +G -2x -G= 5.37
3X= 5.37
X= 5.37 ÷3 <em>(</em><em>÷</em><em>3</em><em> </em><em>on</em><em> </em><em>both</em><em> </em><em>sides</em><em>)</em>
X= 1.79
Subst. X= 1.79 into (1):
2(1.79) +G= 89.61
3.58 +G= 89.61
G= 89.61 -3.58 <em>(</em><em>-3.58</em><em> </em><em>on</em><em> </em><em>both</em><em> </em><em>sides</em><em>)</em>
G= 86.03 <em>(</em><em>simplify</em><em>)</em>
Total bill
= XC +G
= 1.79C +86.03
Thus, the function is B(C)= 1.79C +86.03.