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Question 1 of 5
What are the values of the pronumerals xx and yy?
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First, write the systems of equations being represented in the problem.
55xx+2+2yy |
== |
295295 |
Length |
33xx+4+4yy |
== |
275275 |
Width |
Multiply equation 11 by 22
(5(5xx+2+2yy))×2×2 |
== |
295295×2×2 |
1010xx+4+4yy |
== |
590590 |
Subtract equation 22 from the transformed equation.
1010xx+4+4yy |
== |
590590 |
33xx+4+4yy |
== |
275275 |
|
7x7x |
== |
315315 |
xx |
== |
45 cm45 cm |
Divide both sides by 77 |
33xx+4+4yy |
== |
275275 |
33(45)(45)+4+4yy |
== |
275275 |
Substitute x=45x=45 |
135+4y135+4y-135−135 |
== |
275275-135−135 |
Subtract 135135 from both sides |
4y4y |
== |
140140 |
Divide both sides by 44 |
yy |
== |
35 cm35 cm |
x=45 cmx=45 cm, y=35 cmy=35 cm
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Question 2 of 5
Find the value of the base angles
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First, recall that the base angles of an isosceles triangle are equal
Equate the values of the base angles and simplify
3x-173x−17 |
== |
x+yx+y |
3x-173x−17 -x−x |
== |
x+yx+y -x−x |
Subtract xx from both sides |
2x-172x−17 +17+17 |
== |
yy +17+17 |
Add 1717 to both sides |
2x2x -y−y |
== |
y+17y+17 -y−y |
Subtract yy from both sides |
2x-y2x−y |
== |
1717 |
Next, recall that interior angles of a triangle is equal to 180°180°
Equate the values of the interior angles to 180°180° and simplify
3x-17+x+y+403x−17+x+y+40 |
== |
180180 |
4x+y+234x+y+23 |
== |
180180 |
Combine like terms |
4x+y+234x+y+23 -23−23 |
== |
180180 -23−23 |
Subtract 2323 from both sides |
4x+y4x+y |
= |
157 |
Subtract y from both sides |
Next, write the systems of equations being represented in the problem.
2x-y |
= |
17 |
First equation |
4x+y |
= |
157 |
Second equation |
2x-y |
= |
17 |
4x+y |
= |
157 |
|
6x |
= |
174 |
x |
= |
29 |
Divide both sides by 6 |
2x-y |
= |
17 |
2(29)-y |
= |
17 |
Substitute x=29 |
58-y -58 |
= |
17 -58 |
Subtract 58 from both sides |
-y×(-1) |
= |
-41×(-1) |
Multiply both sides by -1 |
y |
= |
41 |
Finally, substitute the values of x and y to get the value of the base angles.
First equation
x+y |
= |
29+41 |
Substitute known values |
|
= |
70° |
Second equation
3x-17 |
= |
3(29)-17 |
Substitute known values |
|
= |
87-17 |
|
= |
70° |
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Question 3 of 5
A stack of 37 coins consisting of 10 cents and 50 cents adds up to $9.70. How many 10 cent and 50 cent coins are there?
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First, let x be the number of 10 cents coins and y be the number of 50 cents coins
Use these values to create a systems of equations
50x+50y |
= |
970 |
Total value of coins |
x+y |
= |
37 |
Number of coins |
Multiply equation 2 by 10
(x+y)×10 |
= |
37×10 |
10x+10y |
= |
370 |
Subtract the transformed equation from the first equation.
10x+50y |
= |
970 |
10x+10y |
= |
370 |
|
40y |
= |
600 |
y |
= |
15 |
Divide both sides by 40 |
x+y |
= |
37 |
x+15 |
= |
37 |
Substitute y=15 |
x+15 -15 |
= |
37 -15 |
Subtract 15 from both sides |
x |
= |
22 |
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Question 4 of 5
Jason is three times as old as his son Peter. The sum of their ages is 64. What are their ages?
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First, let variables to represent the age of Peter and Jason.
x= Peter’s age
y= Jason’s age
Next, write the systems of equations being represented in the problem.
y |
= |
3x |
Jason is three times as old as Peter. |
x+y |
= |
64 |
The sum of their ages is 64 |
Substitute Equation 1 into 2 and solve for Peter’s age (x).
x+y |
= |
64 |
x+3x |
= |
64 |
y=3x |
4x |
= |
64 |
x |
= |
16 |
Divide both sides by 4 |
Solve for Jason’s age (y).
y |
= |
3x |
y |
= |
3(16) |
Substitute x=16 |
y |
= |
48 |
Simplify |
Peter’s age is 16 while Jason is 48 years old.
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Question 5 of 5
Find the value of x and y
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First, recall that supplementary angles are equal to 180°
Equate the values of the two lower angles to 180° and simplify
3x+2y+3x |
= |
180 |
6x+2y |
= |
180 |
Combine like terms |
Next, recall that alternate angles are equal
Equate the alternate angles and simplify
Next, write the systems of equations being represented in the problem.
6x+2y |
= |
180 |
First equation |
2y |
= |
3x |
Second equation |
Substitute the value of 2y into Equation 1 and solve for x.
6x+2y |
= |
180 |
6x+3x |
= |
180 |
2y=3x |
9x |
= |
180 |
x |
= |
20° |
Divide both sides by 9 |
2y |
= |
3x |
2y |
= |
3(20) |
Substitute x=20 |
2y÷2 |
= |
60÷2 |
Divide both sides by 2 |
y |
= |
30° |