Comprehensive Solutions: Math Homework on Equations and Formulas

This document presents a comprehensive solution to a math homework assignment. The solution covers a variety of algebraic problems including simplifying expressions, solving linear equations, and solving systems of equations. It includes step-by-step solutions to problems involving the ...Read More

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Name:
Course:
Professor:
Date:

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Q1
a) a(5b-2)-5ab
Opening the brackets,
5ab-2a-5ab
Collecting like terms together,
5ab-5ab-2a
= -2a
b) 2a(3b+4)+3b(1-3a)-6b+1
Opening the brackets,
6ab+8a+3b-9ab-6b+1
Collecting like terms together,
6ab-9ab+8a+3b-6b+1
= -3ab+8a-3b+1
Q2
a) 7e-8=-120
7e=-120+8
7e=-112
=e = -112/7 = -16
b) 5(3-2f)=4f-1
= 15-10f = 4f-1
= 15+1=4f+10f
= 14f =16
Therefore f=16/14 = 8/7 = 1 1/7
c) 2(2c+4) = -{(3c+6)/4}
4(4c+8) = -(3c+6)
16c+32 = -3c-6
16c+3c = -6-32

P a g e | 3
19c =-38
C=-2
Q3
a) M=24LT(D-T)
L=6m, D=0.3m, T =5/100 = 0.05m
M = 24 x 6 x 0.05x (0.3-0.05)
= 24x 6 x 0.05 x 0.25
M = 1.8t
b) Transpose and make D subject
M =24LT (D-T)
M/24LT = D-T
D = M/24LT + T
c) Yes, the rigger can safely move the load
The load outside diameter is less than 1m
L=8m, T=0.04m and M=7.5t
Finding the maximum outside diameters D for the load which can be lifted,
7.5= 24 x 8 x 0.04(D-0.04)
D-0.04 = 7.5/(24x8x0.04)
D=0.9765625+0.04
Maximum outside diameter which can be lifted is 1.164625m. this means that the available
diameter of less than 1m will be safely lifted.
Q4
a) i. mobile plan
C= 50 +7/100m, where m is the number of sms sent per month
ii.
total monthly cost = 67.50
67.50 = 50+0.07m
67.50-50=0.07m

P a g e | 4
17.50/0,07 = m
M = 250 sms
b) i.
L= T + 26
ii.
D= 249km
249x2 = 3(sum of L and T)
249x2 = 3(T + T +26)
498=6T+78
iii.
solving for T
498=6T +78
498-78 =6T
T =70km
Q5
Let C represent correct answer, W represent wrong answers and E total earning by the student
Generally E = 10C-7W
C+W= 32
10C-7W =133
C= 32-W
Replacing C in E equation,
10(32-W)-7W = 133
320-10W-7W =133
320-133=17W
187=17W
W=11
C=32-11

P a g e | 5
C=21
Therefore the correct answers were 21
Q6
a) 2d+e=7 and 3d+2e=10
Multiplying equation 1 with 2 and equation 2 with 1
2d+e=7 /2
3d+2e=10/1
We get
4d+2e=14
3d+2e=10
Subtracting the two equations downwards
4d+2e=14
3d+2e=10
D= 4
Replacing value of d in one of the equations; 2(4)+e=7
E=7-8= -1
Therefore d = 4 and e=-1
b) Brand A=$7/kg and B=$11/kg
Let y represent kg in brand A and z kgs in brand B
Therefore; y+z=15
The total cost equation will be 7y+11z=129
Y=15-z
Replacing y in the total cost equation;
7(15-z)+11z=129
105-7z+11z=129
-7z+11z=129-105
4z=24; z=6kg

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Y=15-6 = 9kg
Brand A kgs = 6kg and Brand B = 9kg
Q7
a) i.
y=1/10 x -5
Considering the standard line equation of y=mx+c; where m is the gradient;
Therefore the lines gradient will be 1/10
ii. Coordinates of y intercept means that x=0. Replacing x with 0 in the equation gives y values
as y=1/10x 0 -5. Therefore, y value = -5
Coordinates of y intercept = (0,-5)
b) y intercept of a perpendicular line
The gradient of a perpendicular is usually negative reciprocal
The negative reciprocal of 1/10 = -10. The two points which will be used to achieve this gradient
are (-1,7) and (0,c)
m = y2− y1
x2−x1
-10 = c−7
0−−7
-10(+7) = c-7
-70+7=c = -63
Therefore y intercept coordinates will be (0,-63)
c) gradient
m = y2− y1
x2−x1
m = 18−8
−3−−7 =10
4 = 5
2
d) equation of line
m = y2− y1
x2−x1
m = 1 4−−8
−5−0.5 = 22
−5.5 =−4

P a g e | 7
y=-4x+c
The y intercept will happen at point (0,c), picking one point of the line to find value of c
and using the gradient, we have
-4 = 1 4−c
−5−0
-4x-5 = 14-c
20=14-c
C=-6
The lines equation will be y = -4x-6