Download Evaluate the following expressions

Evaluate the following expressions:
a) (1/3 – ½) – {(-3) – (-4)} -1/6 – 7 = -43/6
(2-3) – (3+4) -1 -7= -8
b)(0.5 + (-0.49)) * (-0.1) – 0.1 / (10*0.1) + 0.1
= 0.01*(-0.1) – 0.1/1 + 0.1
= -0.001
Problem 2.
Evaluate the following expressions:
a) (1/5 – 1/4)^2 – (1/4 -1/5)^2 = (-1/20)2 – (1/20)2 = 0
b) (1/2 – 1/3)^3 – (1/3 – ½)^3 = (1/6)3 – (-1/6)3 = 1/216 + 1/216 = 1/108
Problem 3.
Use fractions (do not convert numbers to their decimal form) to evaluate the expression
x*y + y/z – z/x for the following values of x, y, and z.
a) x =1/2, y = -2/3, z = ¾
1   2    2  4   3 
1 8 3
49
x*y + y/z – z/x = 

     * 2      
2  3   3  3   4 
3 9 2
18
b) x = 10, y = 0, z = -1
x*y + y/z – z/x = 10*0 + 0/(-1) – (-1)/10 = 1/10
Problem 4.
Solve the following equations for x, and explicitly show that your solution is good:
a) 1/2x + 2 = 1/4x + 4
(1/2-1/4)x = 4-2
Or ¼ x = 2
Therefore, x = 8
To show: (1/2)*8 + 2 = 4 + 2 = 6
(1/4)*x + 4 = (1/4)*8 + 4 = 2 + 4 = 6
b) x/a + x/b = b/a + a/b
bx  ax b 2  a 2

ab
ab
b2  a2
Therefore, x =
ab
To show: x/a + x / b =
b2  a2 b2  a2
ba
b2  a2 b a

 (b 2  a 2 )

 
a(a  b) b(a  b)
ab(a  b)
ab
a b
Problem 5.
Solve the following equations for x, and explicitly show that your solution is good:
a) (0.1x – 0.9) * 0 = (10x -10) * 1/10
because left hand side is zero, (10x -10) * 1/10 = 0
Or x = 1
b) 0.25y + x + 1/2x = y + x -0.75y
or x (1 + ½ -1) = y – 0.75y – 0.25y
x=0
Problem 6.
Solve the following inequalities, and write down their solution in the interval notation:
a) -2x<1/2 and x+3/2 > -x
x < -1/4 and 2x > -3/2 or x > -3/4
Therefore, x lies in interval (-3/4, -1/4).
b) (-0.1)^3 x + (-0.5)^2 < 0.999x – 0.75 or –x+1/2 > ¾
-0.001x + 0.25 < 0.999x – 0.75
or 1 < x
or x > 1
–x+1/2 > ¾
Or –x > ¾ - ½ = -1/4
Therefore, x < ¼
Therefore x lies in the interval: (-inf, ¼) union (1, inf)
Problem 7.
Solve the following inequalities, write down their solution in the interval notation:
a) (-x+1) > 2
-x > 2-1 = 1
Therefore, x < -1
Therefore x lies in the interval: (-inf, -1)
b) 1/2x < 1/3 (x+1)
1/2x < 1/3 x + 1/3
or x/6 < 1/3
Therefore, x<2
Therefore x lies in the interval: (-inf, 2)
Problem 8.
The total number of points that you can earn in class is $1000. This counts for 20% (or
200pts) of your grade. So far you earned 90% of the maximum possible number of
points. What is the range of the scores that will result in you earning a grade of B+, if this
grade requires the total number of points N, such that 87% < N < 90%?
The range of score will be 870 < actual score < 900
Problem 9
Write down the slope, the x- and y-intercepts, as well as the equation of line in the slopeintercept form for lines that go through each of the following pairs of points:
a) (1, -1), (-2, 2)
y – (-1) = [(2 +1)/(-2-1)](x-1)
or y + 1 = -x + 1
or y = -x
b) (-1, -2), (3, 4)
y – (-2) = [(4+2)/(3+1)](x+1)
or y + 2 = 2x + 2
or y = 2x
Problem 10
Convert the following equations of line into their slope-intercept form. For each of those
find the equation(also in the slope-intercept form) of a perpendicular line that goes
through the point with coordinates (-1,1):
a) -2x+y = x+2y + 2
y – 2y = x + 2x + 2
Or –y = 3x + 2
Or y = -3x -2
Slope of perpendicular line = -1/-3 = 1/3
Equation of perpendicular line: y = x/3 + c
But this line passes through (-1, 1), therefore, 1 = -1/3 + c or c = 4/3
Equation of perpendicular line: y = x/3 + 4/3
b) x/a + y/b = 1/ab
y/b = -x/a + 1/ab
or y = (-b/a)x + 1/a
Slope of perpendicular line = -1/(-b/a) = 1/b
Equation of perpendicular line: y = ax/b + c
But this line passes through (-1, 1), therefore, 1 = -a/b + c or c = 1+ a/b
Equation of perpendicular line: y = ax/b + a/b