The required answer is 10.5
Answer:
1618 minutes
Step-by-step explanation:
Jerold reads at a rate of 3 pages in 2 minutes. He as a 2427 page novel to read and a 906 page mystery book to read. How long will it take him to read his novel?
From the question above, the number of pages for his novel = 2427 pages
His reading rate =
3 pages = 2 minutes
Hence:
3 pages = 2 minutes
2427 pages = x minutes
Cross Multiply
3 pages × x = 2427pages × 2 minutes
x = 2427pages × 2 minutes/3 pages
x = 1618 minutes
It would take him 1618 minutes to read his 2427 paged novel
Answer: a. 
b. The population in 2032 will be 8292.
Step-by-step explanation:
Exponential growth equation:
, where a = initial value , r = growth rate, x = time period
Given: a = 5054 , r= 0.02
a. Required exponential growth function: 

b. For 2032, x= 25


The population in 2032 will be 8292.
A linear pair is a pair of angles that form a straight line and are therefore, equal to 180 degrees
3n + 19 + 55 + 33 = 180
3n + 107 = 180
3n = 180 - 107
3n = 73
n = 73/3
so < EFG = 3n + 19 = 3(73/3) + 19 = 73 + 19 = 92
and < GFH = 55 + 33 = 88
Step One
======
Find the length of FO (see below)
All of the triangles are equilateral triangles. Label the center as O
FO = FE = sqrt(5) + sqrt(2)
Step Two
======
Drop a perpendicular bisector from O to the midpoint of FE. Label the midpoint as J. Find OJ
Sure the Pythagorean Theorem. Remember that OJ is a perpendicular bisector.
FO^2 = FJ^2 + OJ^2
FO = sqrt(5) + sqrt(2)
FJ = 1/2 [(sqrt(5) + sqrt(2)] \
OJ = ??
[Sqrt(5) + sqrt(2)]^2 = [1/2(sqrt(5) + sqrt(2) ] ^2 + OJ^2
5 + 2 + 2*sqrt(10) = [1/4 (5 + 2 + 2*sqrt(10) + OJ^2
7 + 2sqrt(10) = 1/4 (7 + 2sqrt(10)) + OJ^2 Multiply through by 4
28 + 8* sqrt(10) = 7 + 2sqrt(10) + 4 OJ^2 Subtract 7 + 2sqrt From both sides
21 + 6 sqrt(10) = 4OJ^2 Divide both sides by 4
21/4 + 6/4* sqrt(10) = OJ^2
21/4 + 3/2 * sqrt(10) = OJ^2 Take the square root of both sides.
sqrt OJ^2 = sqrt(21/4 + 3/2 sqrt(10) )
OJ = sqrt(21/4 + 3/2 sqrt(10) )
Step three
find h
h = 2 * OJ
h = 2* sqrt(21/4 + 3/2 sqrt(10) ) <<<<<< answer.