Friday, September 20, 2013
Solutions to Caribbean Algebra II Problems
Graph for the simultaneous solution problem, (4)
Graph for the velocity - time problem, (5)
Solutions to Algebra II test:
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) Factorize completely:
(i) 15 x2 y - 20 xy2
= 5xy (3x - 4y)
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(x + 5)/ 3
3) This graph is easily sketched and is a downward oriented parabola with minimum at (x, y) = (-0.5, -6.2). Inspection of the graph shows that the interval of the domain (first coordinate) for which F(x) < 0 is just: -3 < x < 2
Over this interval the parabola defining the equation is below the zero line for y = F(x)
4) The graph for the quadratic is sketched as the parabola in the uppermost graphic.
a) The gradient for the same curve at the point where x = 2 can be found by carefully placing a ruler or straight edge at the point and bearing in the same inclination. Thereby, one estimates the ratio of the "rise" over the "run". In other words, the change in y coordinates to the change in x coordinates. This will be found to be m = 5. (E.g. roughly 7.5/ 1.5 = 5)
b) The graph of 5y = 18x + 36 is shown as the straight line on the same axes. (To graph this, the student may recast the eqn. as: y = (18/5) x + 36/5.
c) The solutions to the simultaneous equations can be roughly estimated at the intersections of the two graphs. These will then yield the approximate x values to solve both - again very roughly, since eyeballing the values is not a "piece of cake". But the solutions are (approx.) x = -7/6 and x = 4.5.
5) a) The velocity -time graph is shown in the lower graphic at the top of this post.
b) i) The velocity after 2 seconds is read off the graph as v = 12.5 m/s (the acceleration portion of the graph crosses midway between 10 m/s and 15 m/s on the vertical scale.)
ii) The car's acceleration in the first five seconds will the v coordinate maximum (30 m/s) divided by the horizontal or time coordinate (5 s) or: 30 m/s/ 5s = 6 meters per second per second.
iii) The distance travelled in the 15 s is the total area under the v-t graph. In this case, we will add a triangle with base of 5s and altitude 30 m/s to a rectangle with height 30 m/s and width of 10 m/s. Then:
A = 1/2 (30 m/s) (5s) + (30 m/s) (10 s) = 75 m + 300 m = 375 m
iv) The average speed for the 15 seconds is the total distance divided by the time or:
V(av) = D/ t = 375 m/ 15s = 25 m/s
6) a) the key clue is that p varies directly with q, not inversely, so irrespective of how the numbers display in the panels, the proportion: p ~ q is the decider.
Using the values in the table we therefore set up a first proportion to get b:
2/ 8 = 6.1/ b or 2b = 48.8 and b = 24.4
Similarly we can write:
8/ b = a/ 1.2 or 8/ 24.4 = a / (1.2)
Cross multiplying: 8(1.2) = 24.4 a = 9.6
and a = 9.6/ (24.4) or a = 0.4 to the nearest tenth
b) This word problem can easily be solved using two expressions with x denoting pencils and y denoting erasers. Then, we can write:
7x + 5y = 11.60
5x + 3y = 7.60
Solving this system yields: x = 0.80 and y = 1.20
Or, each pencil costs 80 cents and each eraser costs $1.20. Therefore, 8 erasers costs 8 x $1.20 = $9.60.
c) The key clue is that S varies directly as (r + 1)
Then we may write:
s/ 8 = (r + 1)/ 4
when r = 3
s/ 8 = (r + 1)/ 4
Thus, when S = 8, r + 1 = 4 and these are paired (proportionately) in the denominators. The numerators then contain the unknowns and we will set s = 50 to find the value of r corresponding to it:
50/ 8 = (r + 1)/ 4
and: 8 (r + 1) = 4 (50) = 200
Then: 8r + 8 = 200 so: 8r = 200 - 8 = 192
r = 192/ 8 = 24
7) This is a simple application problem. We have a floor in the shape of a rectangle with the length c meters, and the width 2 meters less.
a) i) In terms of c, the width will be: c - 2
ii) The area will be L x W = c (c - 2)
b) If the area of the floor is 15 m2 then the equation in c would be:
c( 2- c) = 15 or: c2 - 2c + 15 = 0
c) The width of the floor can then be determined by factoring the quadratic above, so:
(c - 5) ( c + 3) = 0
SO: c = 5 and c = -3
We dispense with the negative root as unphysical, leaving c = 5 m
Then the width of the floor is (c - 2) = 5m - 2m = 3m