BODMAS
Algebra
Student’s Name
Institution of Affiliation
Question 1
a). i) To work out 6.38–3.560.324+0.412x 2.35
, we use BODMAS formula. The first step is working out the bracket in the denominator and finding the difference in the numerator to obtain;
2.820.133488 x 2.35
Working out the denominator we have;
= 2.820.313697
We divide to obtain;
= 8.99 (3 significant figures). The correct significant figures equal the smallest significant figures of the numbers used in the above expression.
ii). 2π x 3.1232 x (0.74+5.2)
We begin by working out the bracket and the square under the square root to obtain;
=2π x 9.753 x (5.94)
Substituting the value of π =227
and multiplying by the coefficient (2) yields;
= 6.286 x 9.753 x 5.94
We now multiply the values under the square root and express the result to 3 significant figures.
= 364
We find the square root as;
= 19.08 (4 significant figures
iii). 4.32 x 10-322.45 x 103
in this expression, 10-3 has only one significant figure, thus the result of working out the bracket is presented in 1 significant fgure.
= 0.00422450
= 0.000022450
The numerator is expressed in 1 significant figure and the resulting solution is also expressed in 1 significant figure as indicated below;
= 0.000000008 (1 significant figure)
b). 5(4 ÷ 2) + (7 – 4 x 3) + 11. For this expression, we first work out what is in the brackets to get;
= 5(2) + (7- 12) + 11
= 10 – 5 + 11
= 16
c).
i).
The equation is h = h*p1p
……………… (i) Or
p = h1p*h
………………………….. (ii) where p1 is for h* in equation (i) and h is for p* in equation (ii). h1 ≥ h, h* and p* are known.
ii).
h = 24
x 45
Formed by substituting the values of p1, p and h*.
h = 965
h = 19.2 hours
iii).
P = 7.5
x 244
Formed by substituting the values of
= 7.5 x 6
= 45 people
Question 2
a). i). Expanding the expression 3n2 − 2n(3n − 3) − 7 + 4n
= 3n2 – 6n2 + 6n + 4n – 7
= – 3n2 + 10n – 7 multiply all through by -1
= 3n2 – 10n + 7 using the completing square method
= 3n2 – 3n – 7n + 7
= 3n (n – 1) – 7 (n – 1)
= (3n – 7) (n – 1)
ii) -2 (3p – 2) (-2p + 4)
= -2 [3p (-2p + 4) – 2 (-2p + 4)] Expanding the brackets
= -2 [-6p2 + 12p + 4p – 8] Collecting the like terms
= -2 [-6p2 + 16p – 8]
= 12p2 – 32p + 16
b).
i). 32a32 x b223a4b3
= 12a6 b223a4b3
Working out the bracket in the numerator and the known values
= 18a6 b2a4b3
Cancelling the like terms
= 18a2b
or
= 18a2b-1
ii). 12x–3x-3
= x -3– 6x 2x(x-3)
Expanding the brackets
= -5x-3 2x(x-3)
= -5x-3 2x2–
6x
Multiplying both the numerator and the denominator by -1 we get;
= 5x+3 6x-2x2
c). 3 (5 + x) = 7 – 5x
15 + 3x = 7 – 5x
3x + 5x = 7 – 15
8x = 8
x = 1
d).
3x + y = 3 ……………………………………………….. (i)
4x + 3y = -1……………………………………………… (ii)
Making y the subject of equation (i) we have;
y = 3 – 3x ……………… (iii) Substituting the value of y into equation (ii) we get;
4x + 3 (3 – 3x) = -1
4x + 9 – 9x = -1
-5x = -10
x = 2
We substitute the value of x in equation (iii) to obtain the value of y as follows.
y = 3 – 3 (2)
y = 3 – 6
y = -3
e). p=3q+9m2q
………………………………………….. (i)
Multiplying all through by 2q we have;
2pq=6q2+9m
Taking 6q2 to the other side of the equal sign yields;
9m = 2pq – 6q2 Dividing all through by 9s
m= 2pq-6q29
………………………………………….. (ii)
Given that p = 12 and q = 3, we substitute these values in equation (ii) to find the value of m
m= 212(3)-6(3)29
m= 72-549
m= 189
m = 2
Question 3
a).
i).
5a-54=2a+74
Multiplying all through by 4;
20a – 5 = 8a + 7 Collecting the like terms
20a – 8a = 7 + 5
12a = 12 Dividing all through by 12
a = 1
ii)
y2–3y5=15
Multiplying all through by 10 we have;
5y – 6y = 2
-y = 2 Multiplying both sides by -1
y = 2
iii)
4x-3
+ 5 = 8 squaring all through;
4x -3 + 25 = 64 collecting the like terms;
4x = 67 – 25
4x = 42 Dividing all through by 4 we have;
x = 10.5
iv)
m2 + 4 = 40
m2 = 40 – 4
m2 = 36 Taking square root in both sides
m = 6
v)
x+23x=x+12x
Multiplying all through by 6x we have;
2 (x + 2) = 3 (x + 1)
2x + 4 = 3x + 3 Collecting the like terms;
3x – 2x = 4 – 3
x = 1
b).
3x2 – x – 2 = 3x2 – 3x + 2x – 2 we find two numbers that when multiplied yield -6 and when summed yield -1. The two numbers are -3 and 2.
= 3x (x – 1) + 2 (x – 1)
= (3x + 2) (x – 1)
c).
i) (3x + 2)2 – 2 = 9x2 + 8x + 10 Working out the brackets
3x (3x + 2) + 2 (3x + 2) = 9x2 + 8x + 10
9x2 + 6x + 6x + 4 = 9x2 + 8x + 10 Collecting the like terms
9x2 – 9x2 + 12x – 8x = 10 – 4
4x = 6
x = 1.5
ii). 2x2 – 9x + 9 = 0 we find two numbers that when multiplied yield 18 and when summed yield -9. The two numbers are -6 and -3
2x2 – 6x – 3x + 9 = 0
2x (x – 3) – 3(x – 3) = 0
(2x – 3) (x – 3) = 0
Thus, 2x – 3 = 0 or x – 3 = 0
2x = 3 or x = 3
x = 1.5 and/or x = 3
Question 4
a). To convert an angle from degrees to radians, we multiply the degrees provided by π180
. But π = 227
, thus π180=227
x 180=221260=0.0175 (3 SF)
.
Therefore, we multiply the provided angle in degrees by 0.0175. 1150 will be;
1150 = (115 x 0.0175) radian
= 2.007 radian. (3 decimal places)
b). To convert m3 into liters (L), 1 m3 = 1000 L
To convert 15.75 m3 to L, we multiply by 1000 to obtain;
15.75 m3 = (15.75 x 1000) L
= 15750 L
c). The relation between kg/m3 and g/l is that, 1 kg/m3 = 1 g/l.
9550 kg/m3 = (9550 x 1) g/l
= 9550 g/l
d). Given that 1 mi (mile) = 1.6093 km, then dividing both sides by 1.6093 gives;
1 km = 0.6214 mi . The result shows that, 1 km/hr = 0.6214 mi/hr.
But 1 km = 1000 m indicating that 1000 m = 0.6214 mi. dividing both sides by 0.6214 yields;
1 mi = 1609.3 m
Therefore, 1 mi/hr = (1609.33600)
m/s
= 0.447 m/s
Therefore,
25 m/s = 250.447
mi/hr
= 55.93 mi/hr. (4 sf)
Question 5
a). Volume of air in the water tank = Volume of the water tank
= πr2h +23πr3
h = 5 m and r = 2 ms
V = 22722(5)+2322723
= 4407 +35221
= 1320+35221
= 167221
= 79.619 m3 (3 dp).
b).
Surface Area of Water Tank = πr2 + 2πrh + 2πr2
But h = 5 and r = 2
= 22722+
222725+
222722
= 887+4407+1767
= 7047
= 100.57 m2 (2 decimal places)
Collepals.com Plagiarism Free Papers
Are you looking for custom essay writing service or even dissertation writing services? Just request for our write my paper service, and we'll match you with the best essay writer in your subject! With an exceptional team of professional academic experts in a wide range of subjects, we can guarantee you an unrivaled quality of custom-written papers.
Get ZERO PLAGIARISM, HUMAN WRITTEN ESSAYS
Why Hire Collepals.com writers to do your paper?
Quality- We are experienced and have access to ample research materials.
We write plagiarism Free Content
Confidential- We never share or sell your personal information to third parties.
Support-Chat with us today! We are always waiting to answer all your questions.
