Information
You have already completed the quiz before. Hence you can not start it again.
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Loading...
-
Question 1 of 5
Write dd as the subject of the equation below
A=12h(c+d)A=12h(c+d)
Incorrect
Loaded: 0%
Progress: 0%
0:00
Changing the subject means operating on an equation so that a chosen variable remains alone on the left side of the equation
Multiply both sides to 22
| AA |
== |
12h(c+d)12h(c+d) |
|
| AA×2×2 |
== |
12h(c+d)12h(c+d)×2×2 |
|
| 2A2A |
== |
h(c+d)h(c+d) |
12×212×2 cancels out |
| 2A2A |
== |
h(c+d)h(c+d) |
| 2A2A÷h÷h |
== |
h(c+d)h(c+d)÷h÷h |
|
| 2Ah2Ah |
== |
c+dc+d |
h÷hh÷h cancels out |
Finally, subtract cc from both sides
| 2Ah2Ah |
== |
c+dc+d |
|
| 2Ah2Ah -c−c |
== |
c+dc+d -c−c |
|
| 2Ah-c2Ah−c |
== |
dd |
c-cc−c cancels out |
|
| dd |
== |
2Ah-c2Ah−c |
Interchange the sides |
-
Question 2 of 5
Write xx as the subject of the equation below
G(x+y)=H(x-y)G(x+y)=H(x−y)
Incorrect
Loaded: 0%
Progress: 0%
0:00
Changing the subject means operating on an equation so that a chosen variable remains alone on the left side of the equation
| G(x+y)G(x+y) |
== |
H(x-y)H(x−y) |
| Gx+GyGx+Gy |
== |
Hx-HyHx−Hy |
Leave only xx terms on the left
| Gx+GyGx+Gy |
== |
Hx-HyHx−Hy |
| Gx+GyGx+Gy -Hx−Hx |
== |
Hx-HyHx−Hy -Hx−Hx |
Subtract HxHx from both sides |
| Gx+Gy-HxGx+Gy−Hx |
== |
-Hy−Hy |
Hx-HxHx−Hx cancels out |
| Gx+Gy-HxGx+Gy−Hx -Gy−Gy |
== |
-Hy−Hy -Gy−Gy |
Subtract GyGy from both sides |
| Gx-HxGx−Hx |
== |
-Gy-Hy−Gy−Hy |
Gy-GyGy−Gy cancels out |
| Gx-HxGx−Hx |
== |
-Gy-Hy−Gy−Hy |
| x(G-H)x(G−H) |
== |
y(-G-H)y(−G−H) |
| x(G-H)x(G−H) |
== |
-y(G+H)−y(G+H) |
Finally, divide both sides by G-HG−H
| x(G-H)x(G−H) |
== |
-y(G+H)−y(G+H) |
| x(G-H)x(G−H)÷(G-H)÷(G−H) |
== |
-y(G+H)−y(G+H)÷(G-H)÷(G−H) |
|
| xx |
== |
-y(G+H)G-H−y(G+H)G−H |
(G-H)÷(G-H)(G−H)÷(G−H) cancels out |
|
| xx |
== |
y(G+H)H-Gy(G+H)H−G |
Multiply -1−1 to the numerator and denominator |
-
Question 3 of 5
Write xx as the subject of the equation below
1z=1y+1x1z=1y+1x
Incorrect
Loaded: 0%
Progress: 0%
0:00
Changing the subject means operating on an equation so that a chosen variable remains alone on the left side of the equation
Subtract 1y1y from both sides
| 1z1z |
== |
1y+1x1y+1x |
|
| 1z1z -1y−1y |
== |
1y+1x1y+1x -1y−1y |
|
| 1z-1y1z−1y |
== |
1x1x |
1y-1y1y−1y cancels out |
|
| y-zzyy−zzy |
== |
1x1x |
| y-zzyy−zzy |
== |
1x1x |
|
| x(y-z)x(y−z) |
== |
zyzy |
Finally, divide both sides by y-zy−z
| x(y-z)x(y−z) |
== |
zyzy |
| x(y-z)x(y−z)÷(y-z)÷(y−z) |
== |
zyzy÷(y-z)÷(y−z) |
|
| xx |
== |
zyy-zzyy−z |
(y-z)÷(y-z)(y−z)÷(y−z) cancels out |
-
Question 4 of 5
Solve for tt given that v=20v=20, u=8u=8 and a=2a=2
v=u+atv=u+at
Incorrect
Loaded: 0%
Progress: 0%
0:00
Changing the subject means operating on an equation so that a chosen variable remains alone on the left side of the equation
Substitute the given values to solve for tt
| vv |
== |
2020 |
| uu |
== |
88 |
| aa |
== |
22 |
| vv |
== |
u+atu+at |
| 2020 |
== |
8+2t8+2t |
Substitute known values |
| 2020 -8−8 |
== |
8+2t8+2t -8−8 |
Subtract 88 from both sides |
| 1212 |
== |
2t2t |
8-88−8 cancels out |
| 1212÷2÷2 |
== |
2t2t÷2÷2 |
Divide both sides by 22 |
| 66 |
== |
tt |
2÷22÷2 cancels out |
| tt |
== |
66 |
Interchange the sides |
-
Question 5 of 5
Solve for rr given that a=2a=2 and S=25S=25
S=a1-rS=a1−r
Write fractions in the format “a/b”
Incorrect
Loaded: 0%
Progress: 0%
0:00
Changing the subject means operating on an equation so that a chosen variable remains alone on the left side of the equation
Multiply both sides of the equation to 1-r1−r
| SS |
== |
a1-ra1−r |
| SS×(1-r)×(1−r) |
== |
a1-ra1−r×(1-r)×(1−r) |
|
| S(1-r)S(1−r) |
== |
aa |
11-r×(1-r)11−r×(1−r) cancels out |
Divide both sides of the equation to SS
| S(1-r)S(1−r) |
== |
aa |
| S(1-r)S(1−r)÷S÷S |
== |
aa÷S÷S |
|
| 1-r1−r |
== |
aSaS |
Leave only rr terms on the left
| 1-r1−r |
== |
aSaS |
|
| 1-r1−r -1−1 |
== |
aSaS -1−1 |
Subtract 11 from both sides |
|
| -r−r |
== |
aS-1aS−1 |
1-11−1 cancels out |
Next, multiply both sides by -1−1
| -r−r |
== |
aS-1aS−1 |
|
| -r−r×(-1)×(−1) |
== |
(aS-1)(aS−1)×(-1)×(−1) |
|
| rr |
== |
1-aS1−aS |
Finally, substitute the given values to solve for rr
| rr |
== |
1-aS1−aS |
|
| rr |
== |
1-2251−225 |
Substitute known values |
|
| rr |
== |
2525-2252525−225 |
Convert the whole number to a fraction |
|
| rr |
== |
23252325 |
