College Math Homework: Complete Assignment Solutions

This document provides a comprehensive solution to a mathematics homework assignment, covering a range of topics. The solutions include detailed step-by-step explanations for problems involving mixed fractions, algebraic equations, and quadratic equations. The assignment addresses multiple mathematical concepts, including simplifying expressions, solving for variables, and applying formulas to find solutions. The document offers clear and concise methods to solve each problem. The solutions are presented with clear mathematical notations and formulas, making it easier for students to understand and follow the problem-solving process. This resource is designed to help students understand and solve various mathematical problems, providing them with the necessary tools to excel in their studies.

1 Question 1.
1.1
Rewriting Mixed Fraction 22
3 as 11
3 we get
( 2
5 )
1
3 ( 3
4 − 1
4 ) + ( 11
3 )
11
9
Combining 3
4 − 1
4 we get2
4 and factoring out 2 we get1
2 , and we get
= 2
5
1
3 ( 1
2 ) + 11
3
11
9
Removing Parantheses we get
= 2
5
1
3 · 1
2
+ 11
3
11
9
Apply the fraction rule :b
c
a = b
c · a and Divide fractions : a
b
c
d
= a· d
b· c we get
= 2
5· 1
3 · 1
2
+ 11· 9
3· 11
Multiply fractions : a · b
c · d
e = a · b · d
c · e , 2
5
6
+ 3
Apply the fraction rule :a
b
c
= a· c
b we get
= 2· 6
5 + 3
= 12
5 + 3
Convert element to fraction :3 = 3· 5
5
= 3· 5
5 + 12
5
= 3· 5+12
5
= 27
5
1.2
Number of shares left after selling it to her brother and cousin = 1 −1
5 − 4
11 =
24
55
Number of shares sold to Emily =1
2 · 24
55 = 12
55.
1.3
Hence number of shares left =24
55 − 12
55 or 12
55.
1.4
Considering Number of shares was x,12
55x = 72 or x = 330.
1

2 Question 2.
2.1
Rewrite1
4 in power − base form :1
4 = 1
2
2 we get log1
2
1
2
2
and then loga xb = b · loga (x) we get log1
2
1
2
2 = 2 log1
2
1
2
Apply log rule : loga (a) = 1 we get log1
2
1
4 = 2
Similarly we get log3 (27) = 3
Now Apply radical rulen
√ an = a, assuming a ≥ 0
log4
q
(10 + 6)
2 = log4 (16) = 2
log3 (27) − log1
2
1
4 + log4
q
(10 + 6)2 = 3 − 2 + 2 = 3.
2.2
Using 2x = 2x−1+1, 2x+1 = 2x−1+2 and ab+c = abac and Factor out common term 2x−1
we get
2x−1 21 + 22 − 1 = 40
2x−1 = 8
2x−1 = 23
x = 4.
3 Question 3.
3.1
(ab + d)
2 + (a + d) (b − d) = a2b2 + 2abd + d2 + ab − ad + bd − d2
= −ad + 2abd + bd + a2b2 + ab.
3.2
9 a y − 4 d x − 6 d y + 6 a x = 3a(3y + 2x) − 2d(3y + 2x)
= (3a − 2d)(3y + 2x)
3.3
l=20+w
3.4
Area=l x w
Area= w(20+w)
Area= w2 + 20w.
2

3.5
w2 + 20w = 4800
w2 + 20w − 4800 = 0
For a quadratic equation of the form ax2 + bx + c = 0 the solutions are x1, 2 =
−b±√ b2 −4ac
2a
w = −20+√ 202 −4· 1·(−4800)
2· 1
w = −80 or 60
But since Width is positive width=60 and length=80.
4 Question 4.
Let number of boys be x.Then number of girls is 10+4x.Hence x+10+4x=100
5x+10-10=100-10
x = 90
5
x=18.
Hence number of girls is 10+4*18=82.
4.1
2s+5m=41
3s+10m=74.5
Multiplying Equation with -2 we get -4s-10m=-82,
Adding this to equation 2 we get
-s=74.5-82
s = 7.5
Substituting this in equation 1 we get 2*7.5 + 5m = 41
5m=41-15
5m = 26
m=5.20
4.2
8x2 − 2x − 15 = 0
4.3
x2 − 1
4 x − 15
8 = 0
(x − 3
2) ( x = +5
4 ) = 0
Hence x =3
2 , x = −5
4
3

4.4
8x2 − 2x − 15 = 0
x1,2 = −(−2)+
√ (−2) 2 −4· 8(−15)
2· 8
x = 3
2 , x = −5
4
4