Math Essay
Mathematics
Name
Institution
Course
Instructor
Date
Mathematics.
Question 1
(a)
6b2a+2b2a-2b232ab5+8a2b
At the denominator,
8ab(4b4+a2)
Simplifying 4b4+a2 = 2b2+a2b2-a
Therefore,
6b2a+2b2a-2b232ab5+8a2b
=6b2a+2b2a-2b28ab2b2+a2b2-a
=3ba-2b24a2b2-a
=3ba-2b2-4aa-2b2
=3b-4a
(b)
(57- 3)(7+ 43)
57+2021-21-12
=23+1921
(c)
(664x3)x23(4x)14
664x3=(64x3)16
=2x12
2x12x23414x14
Ans=14x-12
(d)
42x+1+35-3x=0
LCM = (2x+1)(5-3x)
Multiply everything by LCM
42x+1(5-3x)+35-3x(2x+1)=0(2x+1)(5-3x)
20-12+6x+3=0
6x=-23
x=-356
(e)
f=πr-gt2r
Multiply everything by 2r
2fr=2πr2-gt
gt=2πr2-2fr
gt=2r(π-f)
g=2rπ-ft
Question 2
A cake decorator has exactly enough icing to cover the top and sides of a round cake that has a diameter of 20 cm and is 8 cm high. It has been decided to change to a cube-shaped cake. What is the largest size of cubic cake that can be iced on the top and sides with this amount of icing? Give this size to the nearest cm, justifying your answer.
Solution;
Surface area =r2+2πrh
(π×102)+(2×π×10×8)
100π+160π
=260π
=816.92cm2
Cube cake
Area =5(l×l)
5l2=816.92
l2=816.925
l2=163.384
l=±12.78
Choosing the positive square root because lengths cannot be negative, the cubic cake can have a side length of at most 13cm2 (to the nearest cm2).
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Suggest three ways in which the well-written solution is better than the poorly-written solution (and so is likely to be awarded considerably higher marks).
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The well-written solution has a flow of working.
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The well-written solution has a good methodology.
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The well-written solution has proper S.I units.
b) Explain where the students went wrong in their calculations, and give the correct answer.
Both students substituted the values wrongly in the formulae.
c) Identify two ways in which this solution differs from Well-written solution 1, and for one of the differences, give a reason why this could be useful.
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Substitution of values. Proper substitution helps a student to acquire correct values in calculation.
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The answer from the first solution was used to acquire the answer in the second solution. Therefore the wrong value was used to calculate the second solution thus getting wrong answers.
Question 3
(a) Show that the point (5, 4) lies on the line with equation 5y = 8x - 20.
Solution;
Substitute x with 5 and y with 4
54= 85- 20
20=40-20
20=20, therefore the points (5, 4) lie on the equation 5y = 8x - 20.
(b) Find the equation of the line through the points (1, 0) and (3, −3).
Solution;
Gradient= yx
(1,0) (3,-3)
-3-33-1=-32
1,0x,yGradient=-32
The equation therefore is:
2y=-3x+3
(d) points of intersection to the nearest integer (2,-2)
(e)
5y=8x-20 (multiply by 2)
2y=-3x+3 (multiply by 5)
10y=16x-40
10y=-15x+15 (subtract)
31x=55
x=5531
x=1.774
Question 4
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Write the quadratic expression 4x2 - 16x - 33 in completed-square form.
ans=x2-2x-334
(ii) Hence solve the equation: 4x2 - 16x - 33.
x2-2x-334=0
x2-2x= 334
x2-2x+222=222+334
x+12=9.25
x+1=9.25
Either, x+1=3.0414
x=2.0414
Or,
x+1=-3.0414
x=-4.0414
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(i) The horizontal distance between Billy and the fence is 3 metres. What will be the exact height of the ball above the ground when it passes directly over the fence?
Solution;
When x=3,
y=0.6+1.2x-0.15x2
y=0.6+1.2(3)-0.15(3)2
y=0.6+3.6-1.35
y=2.85
Therefore, the height of the fence is 2.85m.
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How far away from the fence does the ball hit the ground? Give your solution to one decimal place.
y=0.6+1.2x-0.15x2
when y=0
0=6+12x-1.5x2
0=60+120x-15x2
15x2-120x-60=0
x=-b±b2-4ac2a
x=120±(-120)2-(4×15×-60)2×15
x=120±1800030
x=120±134.164130
Either; x=120+134.164130
x=8.4721
Or;
x=120-134.164130
x=-0.4721
How far away from the fence = 8.4721-3
=5.4721
=5.5m 1d.p
Question 5
2x5+9x4-8x3-49x2+6x+40=0
x2x4+9x3-8x2-49x+6=-40
x=-40
x2x3+9x2-8x-49=-6
x=-6
x2x2+9x-8=49
x=49
2x2+9x-8=0
x=-b±b2-4ac2a
x=-9±92-(4×2×-8)2×2
x=-9±1454
Either;
x=-9+12.04164
x=0.7604
Or;
x=-9-12.04164
x=-5.2604
Values of x= -40, -6, 49, 0.7604, -5.2604