Download Midterm I Solutions

Math B70 Midterm I
Name: _______________
Answers
1. Which of the following could be the graph of a function?
(There may be more than one!)
(A)
(B)
1.
(C)
B, C
(D)
SOLUTION: By the vertical line test (vertical lines cross the graph of a function at most once),
(B) and (C) are both the graphs of functions.
Problems 2 and 3 refer to the function F graphed in the figure:
2. What is the range of F?
2.
{1, 3, 4}
SOLUTION: The range of a function consists of all the y-coordinates of the points on the functions
graph. This function only has three values for its y-coordinates, 1, 3, and 4.
3. What is the domain of F?
3.
[–1, 2]
SOLUTION: The domain of a function consists of all the x-coordinates of the points on the functions
graph. This function only has all values from -1 to 2 for its x-coordinates,
. For this
problem, not including the endpoints of the interval would be OK too, since the graph isn't clear on this
point.
Problems 4 and 5 refer to this function:
4. f (1) = ?
SOLUTION:
4.
5
5.
–1
Since 1 is less than or equal to x=1,
5. f (0) = ?
SOLUTION:
Since x=0 is less than 1,
6. What is the range of the function g(x) = | 5 – x | ?
6.
SOLUTION: Since the output of function g is an absolute value, it cannot be negative. When x=5, is
the input, this function outputs 0. Also this function outputs arbitrarily large numbers (and everything
in between). Thus it's range is all nonnegative real numbers,
.
7. If
f (x) = x + 4
and
g(x) = x2+ 1,
then
( f – g )(2) = ?
7.
1
8.
T=0
SOLUTION:
&
And so:
8. Solve for T:
SOLUTION: First cross multiply (multiply both sides of the equal sign by the least common multiple
of all of its denominators):
Distribute:
We need to get all the terms with T on one side of the equal sign, with all other terms on the opposite
side. To do this, we subtract TS from both sides (moving all T terms to the same side), and also
subtract RS from both sides (moving all non T terms to the opposite side):
Next we factor out the T:
Divide both sides by
, and remember that anything divided by 0 equals 0:
9. Solve for B:
9.
SOLUTION: First multiply both sides of the equal sign by the least common multiple of all of its
denominators, AB. This cancels out all the denominators:
Thus:
10. If Z is inversely proportional to A, and Z = 5.8 when A = 1.3,
find Z when A = 7.0.
(Round off to the nearest tenth)
SOLUTION: “Z is inversely proportional to A” means that
“ Z = 5.8 when A = 1.3”, so
So now we know that
6x + 8y = 7
–6x – 5y = 2
12. 7x – 4 y = 5
1.1
.
, and multiplying both sides by 1.3 allows us to solve for k:
.
Solve each system of linear equations.
11.
10.
Thus when A=7,
For each system, y = ?
11.
SOLUTION: Add the equations, and you get 3y = 9,
SOLUTION: Using Cramer's rule, we get
3
so y = 3.
12.
5x + 2y = 1
So.
13.
x + 2y + z = 1
13.
x – 2y + z = 0
x + 2y – z = 1
SOLUTION: Using Cramer's rule, we get
→
14. You have a jar of 80 coins, consisting of nickels and dimes worth a
total of $5.45. How many dimes are in the jar?
14.
29
SOLUTION: Let n be the number of nickels and d the number of dimes. Then we have the equations:
n + d = 80
5n + 10d = 545
Dividing the second equation by 5 yields the equations:
n + d = 80
n + 2d = 109
Cramer's rule yields
And so,
.
15. Calculate how much 23% acid solution must be mixed with how much 57% solution
to get 12 liters of 40% acid solution. How much if each solution
is needed? NOTE: Set up the equations, but don't solve.
T + F = 12
.23T + .57F = .40(12) = 4.8
15. ________________________
16. Graph the solution set:
x + 5y > 5
5x – y < 5
SOLUTION:
x + 5y > 5 ← (0, 0) is NOT on shaded side of this line.
5x – y < 5 ← (0, 0) is on shaded side of this line.
x + 5y = 5
x | y
0 | 1
5 | 0
5x – y = 5
x | y
0 | –1
1 | 0
17. Solve 9 < 2x + 5 < 11 , and write the solution set in interval notation.
SOLUTION: Subtracting 5 from all 3 sides of the inequality, we get
17.
(2, 3)
For problems 18 and 19, assume A = {a, b, x, y} and B = {a, x, z}.
18.
{a, b, x, y, z}
18.
19.
{a, x}
4 < 2x < 6
Then divide by 2:
2< x <3
A∪B = ?
In interval notation this is (2, 3).
19.
A∩B = ?
20. Find the domain of the function:
20.
SOLUTION: The radicand must be non-negative:
→
21. Find the solution set:
SOLUTION:
| 4 – 5 x | = 14
4 – 5x = 14
–5x = 10
or
21.
4 – 5x = –14
–5x = –18
x = –2
or
x = 18/5
22. Find all solutions, and express in interval notation: | 4 – 5 x | < 14
SOLUTION:
22.
–14 < 4 – 5x < 14
Subtract 4 from each side:
–18 < –5x < 10
Divide by –5:
18/5 >
23. Graph the solution set:
x
> –2
| 4 – 5 x | > 14
SOLUTION:
Subtract 4 from each side:
Divide by –5:
23.
4 – 5x > 14
or
4 – 5x < –14
–5x > 10
or
–5x < –18
x < –2
or
x > 18/5
24. Write using fractional exponent notation:
24.
25. Write using radical notation:
25.
26. Simplify completely:
26.
SOLUTION:
24 divided by 5 equals 4 with remainder 4
18 divided by 5 equals 3 with remainder 3
27. Simplify:
27.
8
SOLUTION:
28. Rationalize the denominator:
28.
SOLUTION:
29. Rationalize the denominator:
29.
SOLUTION:
30. Solve:
30.
x=2
SOLUTION:
Start by squaring both sides:
Subtract 10 and add 3x to both sides:
Factor:
→
or
However, only 2 works when substituted into the original equation. So the only solution is 2.
31. Solve:
SOLUTION:
31.
Start by multiplying both sides by
:
Next, square both sides:
And so:
→
32. Multiply these complex numbers:
SOLUTION:
(2 – i)(1 + 2i)
→
32.
Just FOIL it out:
← remember that
33.
Divide these complex numbers:
33.
SOLUTION:
Multiply the numerator and the denominator by the complex conjugate of the denominator.
Recall that
, so: