The expression of the problem is, 8x.
where "x" stands for the number of dogs. so you would then substitute x with 12. Then you would solve the problem to get your answer.(8•12)
45 pages total
3 on last page
45-3=42
there are 42 on the rest of the pages
on each page, there are 2 more pages in album than then stamps on pages
amount of stamps on pages with equal number of stamps is
A=numberofpgest times number of stamps on pages
number of pages=p
number of stamps per papge=s
a=ps
2 more pages than stamps on pages
p=2+s
total number of stamps per page is s=(45-3)/(p-1)
(what I did is I first got rid of number of stamps on last page, then got rid of the last page)
P=2+s
s=(45-3)/(2+s-1)
s=42/(s+1)
times s+1 both sides
s^2+s=42
minus 42
s^2+s-42=0
factor
(s-6)(s+7)=0
set equal to zero
s-6=0
s=6
s+7=0
s=-7, false, no negative stamps
6 stamps per page
sub
p=2+s
p=2+6
p=8
8 pages
check
last page is 3 so 8-1=7 page left
7*6=42
3+42=45
correct
there are
8 pages in the album
6 stamps per page
Answer:
200-20=180
Step-by-step explanation:
The volume of the trough is V(w) = w³ + 20w² - 429w and the rate of change of the volume over a width of 38 inches to 53 inches is 4695 in³/in
<h3>What is an
equation?</h3>
An equation is an expression that shows the relationship between two or more variables and numbers.
Let w represent the width, hence:
length = w + 33, height = w - 13
Volume (V) = w(w + 33)(w - 13) = w³ + 20w² - 429w
V(w) = w³ + 20w² - 429w
Rate of change = dV/dw = 3w² + 40w - 429
When w = 38, dV/dw = 3(38)² + 40(38) - 429 = 5423
When w = 53, dV/dw = 3(53)² + 40(53) - 429 = 10118
Rate = 10118 - 5423 = 4695 in³/in
The volume of the trough is V(w) = w³ + 20w² - 429w and the rate of change of the volume over a width of 38 inches to 53 inches is 4695 in³/in
Find out more on equation at: brainly.com/question/2972832
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