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Question 1 of 4
Write b as the subject of the equation below
6(a+b)=bx
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Changing the subject means operating on an equation so that a chosen variable remains alone on the left side of the equation
Leave only b terms on the right side
| 6a+6b |
= |
bx |
| 6a+6b -6b |
= |
bx -6b |
Subtract 6b from both sides |
| 6a |
= |
bx-6b |
6b-6b cancels out |
Finally, divide both sides by x-6
| 6a |
= |
b(x-6) |
| 6a÷(x-6) |
= |
b(x-6)÷(x-6) |
|
| 6ax-6 |
= |
b |
(x-6)÷(x-6) cancels out |
|
| b |
= |
6ax-6 |
Interchange the sides |
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Question 2 of 4
Solve for r given that v=400 and h=13
r=√vπh
Use π=3.14
Round your answer to two decimal places
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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 r
| r |
= |
√vπh |
|
|
= |
√400(3.14)×13 |
Substitute known values |
|
|
= |
√40040.82 |
Evaluate |
|
|
= |
√9.8 |
| r |
= |
3.13 |
Rounded to two decimal places |
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Question 3 of 4
Write B as the subject of the equation below
a=√AB
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Changing the subject means operating on an equation so that a chosen variable remains alone on the left side of the equation
Get the square of both sides
| a |
= |
√AB |
|
| a2 |
= |
√AB2 |
|
| a2 |
= |
AB |
| a2 |
= |
AB |
|
| a2×B |
= |
AB×B |
|
| a2B |
= |
A |
1B×B cancels out |
Finally, divide both sides by a2
| a2B |
= |
A |
| a2B÷a2 |
= |
A÷a2 |
|
| B |
= |
Aa2 |
a2÷a2 cancels out |
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Question 4 of 4
Write d as the subject of the equation below
U=a+(n-1)d
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Changing the subject means operating on an equation so that a chosen variable remains alone on the left side of the equation
Subtract a from both sides
| U |
= |
a+(n-1)d |
| U -a |
= |
a+(n-1)d -a |
| U-a |
= |
(n-1)d |
a-a cancels out |
| U-a |
= |
(n-1)d |
| (U-a)÷(n-1) |
= |
(n-1)d÷(n-1) |
|
| U-an-1 |
= |
d |
(n-1)÷(n-1) cancels out |
|
| d |
= |
U-an-1 |
Interchange the sides |
