1) (a) If (3x + 1)/3 - (x - 3)/2 = 2 + (2x - 3)/3
find the value of x:
We begin by simplifying both sides and obtained lowest common denominator:
-> (6x + 2 - 3x + 9)/ 6 = (6 + 2x -3)/3 = 2(2x + 3)/6
-> 6x + 2 - 3x + 9 = 4x + 6
-> 3x + 11 = 4x + 6
-> x = 11 - 6 = 5
Which can be inserted into the original (fractional) equation to verify both sides are equal.
(b)b) Factorize completely:
(i) 15 x^y - 20 xy^2
= 5xy (3 - 4y)
(ii) 3 - 12b^2
= 3 (1 - 4b^2) = 3 (1 - 2b)(1 + 2b)
(c) Given that: m = -3, n = 2, p = -1
find the value of:
m(p - n)^2/ 3p + m
Substituting:
-3(-1 -2)^2/ [3(-1) + (-3)]
= -3(-3)^2/ [-3 + (-3)] = -3(9)/ -6 = 27/6 = 9/2
2) f and g are functions defined as follows:
f: x -> 3x - 5
g: x -> x/2
a) Calculate the value of f(-3)
f(-3) = 3(-3) - 5 = -9 -5 = -14
b) Write expressions for (i) f^-1(x) and (ii) g^-1(x)
(i)the inverse function is obtained by exchanging x and y places and solving for the other variable, viz: x = 3y - 5, then y = (x + 5)/3, therefore:
f^-1(x) = (x + 5)/3
(ii) Similarly, g^-1(x) = 2x (since x = y/2 and hence y = 2x)
c) Hence or otherwise, write an expression for (gf)^-1
This is a composition function of the first two, with the order of priority in operation to the right, e.g. f^-1 first then g^-1. Hence, combining the two:
(gf)^-1 = 2 [(x + 5)/3] = 2/3(x + 5) or 2x/3 + 10
3) The coordinates of the points L and N are (5, 6) and (8, -2), respectively.
(i)State the coordinates of the midpoint M of the line, LN.
This can easily be done by mental estimation, since we know the endpoints of the line. The the midpoint M will be displaced from the L,M positions by half of each of the (x1,y1) coordinates, or: x1 + ½ dx, y1 + ½ dy, where dx = 3 (e.g. 8 - 5), and dy = -8 (e.g. -2 -(6)). . Thence: x + ½ dx = 5 + 1.5 = 6.5 and y2 + ½ dy = 6 + (-8)/2= 6 -4 = 2. Then, M is located at (6.5, 2).
(ii) Calculate the gradient of the line LN
The gradient is defined: (y2 - y1)/ (x2 - x1) = (-2 -6)/(8 - 5) = -8/3
(iii) Determine the equation of the straight line which is perpendicular to LN and which passes through point M.
We already have the coordinates of M, so this straight line must pass through them, hence we can use the point slope form for a straight line, viz.
y - y1 = m (x - x1) to find this line, while realizing that if it is perpendicular to the existing line, then the slope (gradient) must be the inverse, or 3/8 (not -8/3)
By the point slope formula:
y - 2 = 3/8(x - 6.5) = 3/8(x - 13/2)
y - 2 = 3x/8 - 39/16, or:
y = 3x/8 - 39/16 + 32/16 = 3x/8 - 7/16
This line is shown for clarity in the accompanying graph (though it does not have to be given as part of any solution!)
b) An aircraft leaves Jamaica at 13:55 hrs. and travels to Barbados via Antigua. The average speed of the aircraft is 420 km/hr. It arrives in Antigua at 16:45 hrs. local time. Given Antigua is ONE hour AHEAD of Jamaica, compute the distance.
The total time as given is 2 hrs. 50 mins. (16:45 - 13:55), but one must correct for the difference in time zones (1 hr.). Hence, the actual time of flight is:
2 hrs. 50 mins. - 1hr. = 1hr 50 mins.
Then the distance covered will be 420 km (for one hour) plus the extra distance covered in the 50 mins. or 50/60(420 km) = 350 km. Therefore, the distance covered is:
420 km + 350 km = 770 km
(4) (a) Calculate the exact values of:
i) (2.8 + 1.36)/ 4 - 2.7
This is: 3.2
ii) (27/8)^1/3
Here we recognize two cube roots, so:
(27/8)^1/3 = 3/2 = 1.5
b) Calculate 9.72 x 12.05
i) Exactly
This is: 117.126
ii) Correct to two decimal places
This is: 117.12
iii) Correct to 2 significant figures
This is: 110
iv) in standard form
This is 1.17126 x 10^2 (using the exact form translated to standard which is always the number N x 10^n where n is the exponent to which 10 is raised)
5)Using Fig. 1 (previous math blog) and the information therein, calculate (giving reasons):
a) Angle MSQ
The key to the solution lies in recognizing the presence of the similar triangles: SRM (to SRP), and SQR (to SPQ). Once these are in place, and also using the results for supplemental angles one can find all the angles given. Thus angle MSQ = 24 deg.
b)Angle RSP
By similar triangles (SQR and SPQ): < RSP = (78 - 26)deg = 52 deg
c)Angle SPN
By similar triangles(SMQ and PQN), alternate interior angles being equal, and working therefrom: angle SPN = 102 deg.
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