Transcript ELEMENTARY-ALG SAMPLE TEST DOWNLOAD

This is a sample test for the
elementary algebra diagnostic test
at the Harbor College.
This is an interactive PowerPoint file.
For the best viewing, download it and run it in
the viewing–mode.
The solution and a link to the review of the
topics are provided for each question.
The review–links of all the topics may be
http://www.lahc.edu/math/algebra/ele-algaccessed at here
place-review.html
Question 1. A bag contained 48 pieces of
candies. Joe took 2/3 of the bag,
Mary took 3/4 of what was left,
A 8
and Chuck got the rest.
How many pieces of candies did
B 6
Chuck get?
C
4
D
3
Answer for Question 1:
Solution to Q1
The correct answer is 4.
The statement “(fraction or %) of an amount”
is translated to the multiplication operation.
Solution to Q1
The correct answer is 4.
The statement “(fraction or %) of an amount”
is translated to the multiplication operation.
Joe took ¾ of the bag so ¼ of the bag was left,
12
1
i.e. 4 (48) = 12 pieces were left.
Solution to Q1
The correct answer is 4.
The statement “(fraction or %) of an amount”
is translated to the multiplication operation.
Joe took ¾ of the bag so ¼ of the bag was left,
12
1
i.e. 4 (48) = 12 pieces were left.
Mary took 2/3 so Chuck got 1/3 of 12,
4
1
i.e. 3 (12) = 4 pieces.
Link to review
Next Question
Question 1. A bag contained 48 pieces of
candies. Joe took 2/3 of the bag,
Mary took 3/4 of what was left,
A 8
and Chuck got the rest.
How many pieces of candies did
B 6
Chuck get?
C
4
D
3
Correct!
Next Question
Question 2. 1 2
3
1
A –1 3
B
2
–1 3
C
1
–2
3
D
2
–2 3
31 =
3
Answer for Question 2:
Solution to Q2
2
The correct answer is –1 3 .
1
2
3
13
3
2
1
1
= 33
3
Solution to Q2
2
The correct answer is –1 3 .
1
2
3
13
3
2
1
1
= 33
3
2
1
=
2
3
3
2
1
=
3
hence 1 2
3
2
1
3
= –1 3
3
Link to review
Next Question
Question 2. 1 2
3
1
A –1 3
B
2
–1 3
C
1
–2
3
D
2
–2 3
31 =
3
Correct!
Next Question
Question 3. After Joe got a 8% raise he earns
$1,350 per week. How much was his weekly
wage before the raise?
A
$1,250
B
$1,225
C
$1,200
D
$1,175
Answer for Question 3:
Solution to Q3
The answer is $1,250.
Suppose Joe’s wage was x before the raise,
then a raise of 8% is 0.08x.
Solution to Q3
The answer is $1,250.
Suppose Joe’s wage was x before the raise,
then a raise of 8% is 0.08x.
Hence his new wage is x + 0.08x or 1.08x,
which is $1,350.
Solution to Q3
The answer is $1,250.
Suppose Joe’s wage was x before the raise,
then a raise of 8% is 0.08x.
Hence his new wage is x + 0.08x or 1.08x,
which is $1,350.
Therefore 1.08x = 1350 so that
x = 1350 = 1250.
1.08
Link to review
Next Question
Question 3. After Joe got a 8% raise he earns
$1,350 per week. How much was his weekly
wage before the raise?
A
$1,250
B
$1,225
C
$1,200
D
$1,175
Correct!
Next Question
Question 4.
4,000 x (7.5 x 10–9) x (8 x 103) =
A
240.
B
24
C
2.4
D
0.24
Answer for Question 4:
Solution to Q4
The answer is 0.24.
4,000 x (7.5 x 10–9) x (8 x 103)
= 4 x 103 x (7.5 x 10–9) x (8 x 103)
Solution to Q4
The answer is 0.24.
4,000 x (7.5 x 10–9) x (8 x 103)
= 4 x 103 x (7.5 x 10–9) x (8 x 103)
= 4 x 7.5 x 8 x 103 x 10–9 x 103
Next Question
Solution to Q4
The answer is 0.24.
4,000 x (7.5 x 10–9) x (8 x 103)
= 4 x 103 x (7.5 x 10–9) x (8 x 103)
= 4 x 7.5 x 8 x 103 x 10–9 x 103
= 240 x 10-3
= 0.24
Link to review
Next Question
Question 4.
4,000 x (7.5 x 10–9) x (8 x 103) =
A
240.
B
24
C
2.4
D
0.24
Correct!
Next Question
Question 5. Given the right triangle as shown
with c = 12, a = 5, which is the closest
approximated value of x?
A
12
B
11
C
10
D
9
Answer for Question 5:
Solution to Q5
The answer is 11.
By the Pythagorean Theorem it must be that
x2 + 52 = 122
Solution to Q5
The answer is 11.
By the Pythagorean Theorem it must be that
x2 + 52 = 122
x2 + 25 = 144
x2 = 119 or that
x = √119 ≈ 11
Link to review
Next Question
Question 5. Given the right triangle as shown
with c = 12, a = 5, which is the closest
approximated value of x?
A
12
B
11
C
10
D
9
Correct!
Next Question
Question 6. Which of the following
percentages is closest to 2/7?
A
30%
B
25%
C
20%
D
15%
Answer for Question 6:
Solution to Q6
The answer is 30%.
2/7 = 0.285.. ≈ 30%
Link to review
Next Question
Question 6. Which of the following
percentages is closest to 2/7?
A
30%
B
25%
C
20%
D
15%
Correct!
Next Question
Question 7.
4.5 x 0.2 – (3.5 – 2 x 0.85) =
A
7.2
B
–0.9
C
10
D
–4.8
Answer for Question 7:
Solution to Q7
The answer is 7.2.
4.5 x 0.2 – (3.5 – 2 x 0.85)
= 0.9 – (3.5 – 1.7)
Solution to Q7
The answer is 7.2.
4.5 x 0.2 – (3.5 – 2 x 0.85)
= 0.9 – (3.5 – 1.7)
= 0.9 – 1.8
= –0.9
Link to review
Next Question
Question 7.
4.5 x 0.2 – (3.5 – 2 x 0.85) =
A
7.2
B
–0.9
C
10
D
–4.8
Correct!
Next Question
3
Question 8. 3 15 – 26 is approximately
A
13
B
11
C
9
D
7
Answer for Question 8:
Solution to Q8
The answer is 9.
3 15 is approximately 3(4) = 12,
and
3
26 is approximately 3
Next Question
Solution to Q8
The answer is 9.
3 15 is approximately 3(4) = 12,
and
3
26 is approximately 3 so that
3
3 15 – 26
≈ 12 – 3 = 9.
Link to review
3
Question 8. 3 15 – 26 is approximately
A
13
B
11
C
9
D
7
Correct!
Next Question
Question 9. If x = 3, and that xy – x – y = –4,
what is y?
A
3/2
B
–3/2
C
1/2
D
–1/2
Answer for Question 9:
Solution to Q9
The answer is -1/2.
If x = 3, then xy – x – y = –4
is 3y – 3 – y = –4.
Solution to Q9
The answer is -1/2.
If x = 3, then xy – x – y = –4
is 3y – 3 – y = –4 so that
2y = –1
or
y = –1/2
Link to review
Next Question
Question 9. If x = 3, and that xy – x – y = –4,
what is y?
A
3/2
B
–3/2
C
1/2
D
–1/2
Correct!
Next Question
Question 10. If x = –3, then –x2 – 2x + 3 is
A
0
B
6
C
12
D
18
Answer for Question 10:
Solution to Q10
The answer is 8.
If x = –3, then –x2 – 2x + 3 is
–(–3)2 – 2(–3) + 3
= –9 + 6 + 3
=0
Link to review
Next Question
Question 10. If x = –3, then –x2 – 2x + 3 is
A
0
B
6
C
12
D
18
Correct!
Next Question
Question 11. If a = –2, b = –1, c = 3,
then b2 – 4ac is
A
23
B
25
C
–25
D
–23
Answer for Question 11:
Solution to Q11
The answer is 25.
If a = –2, b = –1, c = 3,
then (–1)2 – 4(–2)(3)
= 1 + 24
= 25
Link to review
Next Question
Question 11. If a = –2, b = –1, c = 3,
then b2 – 4ac is
A
23
B
25
C
–25
D
–23
Correct!
Next Question
Question 12. If a – bx = c, then x =
A
𝑎
+b
𝑐
B
𝑐−𝑎+𝑏
C
𝑐
𝑎– 𝑏
D
𝑎−𝑐
𝑏
Answer for Question 12:
Solution to Q12
𝑎−𝑐
The answer is 𝑏 .
If a – bx = c
then a – c = bx
so
𝑎−𝑐
=x
𝑏
Link to review
Next Question
Question 12. If a – bx = c, then x =
A
𝑎
+b
𝑐
B
𝑐−𝑎+𝑏
C
𝑐
𝑎– 𝑏
D
𝑎−𝑐
𝑏
Correct!
Next Question
Question 13. –2(3A – 4B) – 5(–7A + 6B) =
A
29A – 22B
B
–41A – 22B
C
29A – 38B
D
–41A – 38B
Answer for Question 13:
Solution to Q13
The answer is 29A –22B.
–2(3A – 4B) – 5(–7A + 6B)
= –6A + 8B + 35A – 30B
= 29A – 22B
Link to review
Next Question
Question 13. –2(3A – 4B) – 5(–7A + 6B) =
A
29A – 22B
B
–41A – 22B
C
29A – 38B
D
–41A – 38B
Correct!
Next Question
14. Given the system
2x – y = 1
3x – y = –1
the solution for x is:
A
2
B
0
C
–2
D
–4
Answer for Question 14:
Solution to Q14
The answer is x = –2.
Subtract the two equations
–)
3x – y = –1
2x – y = 1
x
= –2
Link to review
Next Question
14. Given the system
2x – y = 1
3x – y = –1
the solution for x is:
A
2
B
0
C
–2
D
–4
Correct!
Next Question
Question 15.
A
x y3
B
y3/x
C
x5 y3
D
𝑥 2 𝑦 –7
𝑥 –3 𝑦 –4
=
x5/y3
Answer for Question 15:
Solution to Q15
The answer is x5/y3
By the Divide–Subtract rule for the exponents:
𝑥 2 𝑦 –7
𝑥 –3 𝑦 –4
= x 2–(–3) y –7–(–4)
Solution to Q15
The answer is x5/y3
By the Divide–Subtract rule for the exponents:
𝑥 2 𝑦 –7
𝑥 –3 𝑦 –4
= x 2–(–3) y –7–(–4)
= x5y–3
= x5/y3
Link to review
Next Question
Question 15.
A
x y3
B
y3/x
C
x5 y3
D
x5/y3
𝑥 2 𝑦 –7
𝑥 –3 𝑦 –4
=
Correct!
Next Question
Question 16. From experience, it’s sunny
5 out of 7 days at Sunny Hills. In a period of
182 days, how many cloudy days are
expected?
A
130 days
B
125 days
C 120 days
D
115 days
Answer for Question 16:
Solution to Q16
The answer is 130 days.
Let x be the number of sunny days,
using proportion,
The number of sunny days
Total number of days
x
5
=
182 7
Solution to Q16
The answer is 130 days.
Let x be the number of sunny days,
using proportion,
x
The number of sunny days
5
=
Total number of days
182 7
cross–multiplying so that 7x = 182(5)
Solution to Q16
The answer is 130 days.
Let x be the number of sunny days,
using proportion,
x
The number of sunny days
5
=
Total number of days
182 7
cross–multiplying so that 7x = 182(5)
26
182(5)
x=
7
x = 130
Link to review
Next Question
Question 16. From experience, it’s sunny
5 out of 7 days at Sunny Hills. In a period of
182 days, how many cloudy days are
expected?
A
130 days
B
125 days
Correct!
C 120 days
D
115 days
Next Question
Question 17. y is inversely proportional to x2,
if x = 2 then y = 9, what is x if y = 4?
A
6
B
4
C
3
D
2
Answer for Question 17:
Solution to Q17
The answer is 3.
y is inversely proportional to
x2
k
means y = x2
Solution to Q17
The answer is 3.
k
y is inversely proportional to means y = x2
k
if x = 2, y = 9 then 9 = 2 so k = 36,
2
36
and that y = 2
x
x2
Solution to Q17
The answer is 3.
k
y is inversely proportional to means y = x2
k
if x = 2, y = 9 then 9 = 2 so k = 36,
2
36
and that y = 2
x
36
so if y = 4, then 4 = 2
x
x2
Solution to Q17
The answer is 3.
k
y is inversely proportional to means y = x2
k
if x = 2, y = 9 then 9 = 2 so k = 36,
2
36
and that y = 2
x
36
so if y = 4, then 4 = 2
x
4x2 = 36
x2 = 9 and that x = 3 or –3.
Link to review
x2
Next Question
Question 17. y is inversely proportional to x2,
if x = 2 then y = 9, what is x if y = 4?
A
6
B
4
C
3
D
2
Correct!
Next Question
Question 18. (x – 2)(2x – 3) – (3x + 2)(x – 4) =
A
–2x2 + 3x – 2
B
–2x2 + 3x + 14
C –2x2 – 17x – 2
D –2x2 – 3x + 14
Answer for Question 18:
Solution to Q18
The answer is –2x2 + 3x + 14.
(x – 2)(2x – 3) – (3x + 2)(x – 4)
= (x – 2)(2x – 3) + (–3x – 2)(x – 4)
Solution to Q18
The answer is –2x2 + 3x + 14.
(x – 2)(2x – 3) – (3x + 2)(x – 4)
= (x – 2)(2x – 3) + (–3x – 2)(x – 4)
= x2 – 7x + 6 – 3x2 + 10x + 8
= –2x2 + 3x + 14
Link to review
Next Question
Question 18. (x – 2)(2x – 3) – (3x + 2)(x – 4) =
A
–2x2 + 3x – 2
B
–2x2 + 3x + 14
Correct!
C –2x2 – 17x – 2
D –2x2 – 3x + 14
Next Question
Question 19. Starting with a full tank of gas,
we drove at a constant speed for 2½ hr and
we used up 2/5 of the tank. How much more
time we can travel at the same speed before
we run out of gas?
A
4 hr
B
3¾ hr
C
3½ hr
D
3¼ hr
Answer for Question 19:
Solution to Q19
The answer is 3 43 hrs.
Use proportion to solve the problem.
Let x be the number of hours that we may
travel on the remaining 3/5 tank of gas.
Solution to Q19
The answer is 3 43 hrs.
Use proportion to solve the problem.
Let x be the number of hours that we may
travel on the remaining 3/5 tank of gas.
Hours of traveling time:
Fraction of the tank:
x
5/2
=
2/5
3/5
Solution to Q19
The answer is 3 43 hrs.
Use proportion to solve the problem.
Let x be the number of hours that we may
travel on the remaining 3/5 tank of gas.
Hours of traveling time:
Fraction of the tank:
x
5/2
=
2/5
3/5
2x = 3 5
5
5 2
cross–multiply
Solution to Q19
The answer is 3 43 hrs.
Use proportion to solve the problem.
Let x be the number of hours that we may
travel on the remaining 3/5 tank of gas.
Hours of traveling time:
Fraction of the tank:
x
5/2 cross–multiply
=
2/5
3/5
2x = 3 5 = 3
5
5 2 2
Solution to Q19
The answer is 3 43 hrs.
Use proportion to solve the problem.
Let x be the number of hours that we may
travel on the remaining 3/5 tank of gas.
Hours of traveling time:
Fraction of the tank:
x
5/2 cross–multiply
=
2/5
3/5
2x = 3 5 = 3
5
5 2 2
Solution to Q19
The answer is 3 43 hrs.
Use proportion to solve the problem.
Let x be the number of hours that we may
travel on the remaining 3/5 tank of gas,
Hours of traveling time:
Fraction of the tank:
x
5/2 cross–multiply
=
2/5
3/5
2x = 3 5 = 3 cross again
5
5 2 2
4x = 15
Link to review
x = 15/4
Next Question
= 3 34
Question 19. Starting with a full tank of gas,
we drove at a constant speed for 2½ hr and
we used up 2/5 of the tank. How much more
time we can travel at the same speed before
we run out of gas?
A
4 hr
B
3¾ hr
C
3½ hr
D
3¼ hr
Correct!
Next Question
Question 20. Which of the following is a factor
of 3x – 3y + ax – ay
A
3–a
B
a–3
C
y+x
D
x–y
Answer for Question 20:
Solution to Q20
The answer is (x – y)
Factoring by grouping:
3x – 3y + ax – ay
= 3(x – y) + a(x – y)
Solution to Q20
The answer is (x – y)
Factoring by grouping:
3x – 3y + ax – ay
= 3(x – y) + a(x – y)
= (x – y) (3 + a)
Link to review
Next Question
Question 20. Which of the following is a factor
of 3x – 3y + ax – ay
A
3–a
B
a–3
C
y+x
D
x–y
Correct!
Next Question
Question 21.
–1 – x–1
Combine and simplify 2x
x–3
x+2
A
x2 + 7x + 1
(x – 3)(x + 2)
B
x2 – x – 5
(x – 3)(x + 2)
C
x2 – x + 1
(x – 3)(x + 2)
D
x2 + 7x – 5
(x – 3)(x + 2)
Answer for Question 21:
Solution to Q21
2 + 7x – 5
x
The answer is
(x – 3)(x + 2)
Multiply the expression by 1 in the form of
LCD/LCD distribute and simplify:
2x – 1 – x – 1 (x – 3)(x + 2)
(x – 3)(x + 2)
x–3
x+2
Solution to Q21
2 + 7x – 5
x
The answer is
(x – 3)(x + 2)
Multiply the expression by 1 in the form of
LCD/LCD distribute and simplify:
(x + 2)
(x – 3)
2x – 1 – x – 1 (x – 3)(x + 2)
(x – 3)(x + 2)
x–3
x+2
Solution to Q21
2 + 7x – 5
x
The answer is
(x – 3)(x + 2)
Multiply the expression by 1 in the form of
LCD/LCD distribute and simplify:
(x + 2)
(x – 3)
2x – 1 – x – 1 (x – 3)(x + 2)
(x – 3)(x + 2)
x–3
x+2
= (2x – 1) (x + 2) – (x – 1)(x – 3)
(x – 3)(x + 2)
2 + 7x – 5
x
=
(x – 3)(x + 2)
Link to review
Next Question
Question 21.
–1 – x–1
Combine and simplify 2x
x–3
x+2
A
x2 + 7x + 1
(x – 3)(x + 2)
B
x2 – x – 5
(x – 3)(x + 2)
C
x2 – x + 1
(x – 3)(x + 2)
D
x2 + 7x – 5
(x – 3)(x + 2)
Correct!
Next Question
Question 22. The solutions of
3x(x – 1) = x2 + 9 are
A
x = 3/2, –2
B
x = –3/2, 2
C
x = 2/3, –3
D
x = –2/3, 3
Answer for Question 22:
Solution to Q22
The answer is x = –3/2, 3
3x(x – 1) = x2 + 9
3x2 – 3x = x2 + 9
2x2 – 3x – 9 = 0
Solution to Q22
The answer is x = –3/2, 3
3x(x – 1) = x2 + 9
3x2 – 3x = x2 + 9
2x2 – 3x – 9 = 0
(2x + 3)(x – 3 ) = 0
so x = –3/2, 3
Link to review
Next Question
Question 22. The solutions of
3x(x – 1) = x2 + 9 are
A
x = 3/2, –2
B
x = –3/2, 2
C
x = 2/3, –3
D
x = –2/3, 3
Correct!
Next Question
Question 23. Solve for x if a = 1 – 1
x
A
B
C
D
1
1–a
1
1+a
–1
1+a
1
a–1
Answer for Question 23:
Solution to Q23
1
The answer is x =
1–a
a=1– 1
x
1 =1–a
x
1
x=
1–a
Link to review
Next Question
Question 23. Solve for x if a = 1 – 1
x
A
B
C
D
1
1–a
1
1+a
–1
1+a
1
a–1
Correct!
Next Question
Question 24.
(1 – x2) (x2 – 3x)
Simplify
x2 * (x2 – 4x + 3)
A
x+1
x
B
–x – 1
x
C
x–1
x
D
1–x
x
Answer for Question 24:
Solution to Q24
The answer is –xx– 1
(1 – x2) (x2 – 3x)
x2 * (x2 – 4x + 3)
(1 – x)(1 + x) x(x – 3)
=
x2(x – 3)(x – 1)
Solution to Q24
The answer is –xx– 1
(1 – x2) (x2 – 3x)
x2 * (x2 – 4x + 3)
–1
(1 – x)(1 + x) x(x – 3)
=
x2(x – 3)(x – 1)
–x – 1
=
x
Link to review
Next Question
Question 24.
(1 – x2) (x2 – 3x)
Simplify
x2 * (x2 – 4x + 3)
A
x+1
x
B
–x – 1
x
C
x–1
x
D
1–x
x
Correct!
Next Question
Question 25. The vertices of a rectangle are
(1, 1), (5, 1), (1, 7), and (5, 7).
What is the area of the rectangle?
A
16
B
20
C
24
D
28
Answer for Question 25:
Solution to Q25
The answer is 24.
We have
(1, 5)
(7, 5)
Δy = 5 – 1 = 4
(1, 1)
Δx = 7 – 1 = 6
(7, 1)
So the area of the rectangle is
Link to review
Δx Δy = 6×4 = 24.
Next Question
Question 25. The vertices of a rectangle are
(1, 1), (5, 1), (1, 7), and (5, 7).
What is the area of the rectangle?
A
16
B
20
C
24
D
28
Correct!
Next Question
Question 26. The solution for the inequality
3 > 1 – 2x ≥ –3 is
A
–1
2
B
–1
2
C
–1
2
D
–1
2
Answer for Question 26:
Question 26.
The answer is
3 > 1 – 2x ≥ –3
2 > – 2x ≥ –4
–1 < x ≤ 2 or
–1
2
subtract 1
divide by –2
and reverse the inequality
–1
2
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Next Question
Question 26. The solution for the inequality
3 > 1 – 2x ≥ –3 is
A
–1
2
B
–1
2
C
–1
2
D
–1
Correct!
2
Next Question
Question 27. Following are two prints of
different sizes of the same picture of Tomo.
With the given measurements, what is x?
A
2
2 ft
3
2 ft
x
B
3 ft
3 ft
5 ft
C
2
3 ft
3
D
1
3 ft
3
Answer for Question 27:
Solution to Q27
The answer is 3 1/3 ft.
The ratios of the measurements must satisfies
the proportion x : 5 = 2 : 3 or that
x = 2
5
3
2 ft
x
3 ft
5 ft
Next Question
Solution to Q27
The answer is 3 1/3 ft.
The ratios of the measurements must satisfies
the proportion x : 5 = 2 : 3 or that
x = 2 so
5
3
1
2
x = (5)= 3 3
3
2 ft
x
3 ft
5 ft
Link to review
Next Question
Question 27. Following are two prints of
different sizes of the same picture of Tomo.
With the given measurements, what is x?
A
2
2 ft
3
2 ft
x
B
3 ft
3 ft
5 ft
C
2
3 ft
3
D
1
3 ft
3
Correct!
Next Question
Question 28.
Which of the following pictures represents
A, B and C on the real line most accurately
if A = 3, B = 9 and C = 12?
A
0
B
0
C
0
D
0
Answer for Question 28:
Solution to Q28
The answer B.
C = 12 must be the right most point.
0
12
Solution to Q28
The answer B.
C = 12 must be the right most point.
By dividing from 0 to 12 into 4 pieces
0
12
Solution to Q28
The answer B.
C = 12 must be the right most point.
By dividing from 0 to 12 into 4 pieces
we see that A = 3, B = 9 and C = 12 must be
0
3
6
9
12
Link to review
Next Question
Question 28.
Which of the following pictures represents
A, B and C on the real line most accurately
if A = 3, B = 9 and C = 12?
A
0
Correct!
B
0
C
0
D
0
Next Question
Question 29. Let point A = (–10, 20),
the coordinate of the point that is 40 to the right
and 40 below A is
A
(40, –30)
B
(30, –30)
C
(40, –20)
D
(30, –20)
Answer for Question 29:
Solution to Q29
The answer (30, –20).
the point that’s 40 to the right and 40 below
of (–10, 20) is (–10 + 40, 20 – 40) = (30, –20)
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Next Question
Question 29. Let point A = (–10, 20),
the coordinate of the point that is 40 to the right
and 40 below A is
A
(40, –30)
B
(30, –30)
C
(40, –20)
D
(30, –20)
Correct!
Next Question
Question 30. The slope of the line that has
x–intercept at 4 and y intercept at 3 is
A
3/4
B
–3/4
C
4/3
D
–4/3
Answer for Question 30:
Solution to Q30
The answer is –3/4.
The slope is “rise/run”.
Solution to Q30
The answer is –3/4.
The slope is “rise/run”.
The rise and run for (4, 0) and (0, 3) is
(4, 0)
(0, 3)
4,–3
run
rise
Hence the slope is –3/4.
Link to review
Next Question
Question 30. The slope of the line that has
x–intercept at 4 and y intercept at 3 is
A
3/4
B
–3/4
C
4/3
D
–4/3
Correct!
Next Question
Question 31. The slope of a vertical line is
A
Undefined
B
0
C
1
D
–1
Answer for Question 31:
Solution to Q31
The answer is “undefined”.
The slope is “rise/run”.
The run for two points on a vertical line is 0
hence the slope = rise/0 is not defined.
Link to review
Next Question
Question 31. The slope of a vertical line is
A
Undefined
B
0
C
1
D
–1
Correct!
Next Question
Question 32. Find the y–intercept of the line
that contains the points (4, 15) and (9, 45).
A
–8
B
–9
C
–10
D
–11
Answer for Question 32:
Solution to Q32
The answer is –9.
The slope of the line that contains the points
(4, 15) and (9, 45) is (45 –15)/(9 – 40) = 6
Solution to Q32
The answer is –9.
The slope of the line that contains the points
(4, 15) and (9, 45) is (45 –15)/(9 – 40) = 6
so the equation of the line containing these
points is y = 6(x – 4) + 15
Solution to Q32
The answer is –9.
The slope of the line that contains the points
(4, 15) and (9, 45) is (45 –15)/(9 – 40) = 6
so the equation of the line containing these
points is y = 6(x – 4) + 15 or
y = 6x – 9
Hence the y–intercept is at y = –9.
Link to review
Question 32. Find the y–intercept of the line
that contains the points (4, 15) and (9, 45).
A
–8
B
–9
C
–10
D
–11
Correct!