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Question 1 of 5
Write rr as the subject of the equation below
V=πr2hV=πr2h
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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
Divide both sides of the equation by πhπh
VV |
== |
πr2hπr2h |
VV÷πh÷πh |
== |
πr2hπr2h÷πh÷πh |
|
VπhVπh |
== |
r2r2 |
πh÷πhπh÷πh cancels out |
Get the square root of both sides
VπhVπh |
== |
r2r2 |
|
√Vπh√Vπh |
== |
√r2√r2 |
|
√Vπh√Vπh |
== |
rr |
|
rr |
== |
√Vπh√Vπh |
Interchange the sides |
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Question 2 of 5
Write aa as the subject of the equation below
b=a-3a-4b=a−3a−4
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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
Multiply both sides of the equation to a-4a−4
bb |
== |
a-3a-4a−3a−4 |
bb×(a-4)×(a−4) |
== |
a-3a-4a−3a−4×(a-4)×(a−4) |
|
b(a-4)b(a−4) |
== |
a-3a−3 |
1a-4×(a-4)1a−4×(a−4) cancels out |
|
ab-4bab−4b |
== |
a-3a−3 |
Leave only aa terms on the left
ab-4bab−4b |
== |
a-3a−3 |
ab-4bab−4b -a−a |
== |
a-3a−3 -a−a |
Subtract aa from both sides |
ab-4b-aab−4b−a |
== |
-3−3 |
a-aa−a cancels out |
ab-4b-aab−4b−a +4b+4b |
== |
-3−3 +4b+4b |
Add 4b4b to both sides |
ab-aab−a |
== |
4b-34b−3 |
-4b+4b−4b+4b cancels out |
ab-aab−a |
== |
4b-34b−3 |
a(b-1)a(b−1) |
== |
4b-34b−3 |
Finally, divide both sides by b-1b−1
a(b-1)a(b−1) |
== |
4b-34b−3 |
a(b-1)a(b−1)÷(b-1)÷(b−1) |
== |
(4b-3)(4b−3)÷(b-1)÷(b−1) |
|
aa |
== |
4b-3b-14b−3b−1 |
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Question 3 of 5
Write xx as the subject of the equation below
7-3x=mx+t7−3x=mx+t
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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
Keep only xx terms on the right side
7-3x7−3x |
== |
mx+tmx+t |
7-3x7−3x +3x+3x |
== |
mx+tmx+t +3x+3x |
Add 3x3x to both sides |
77 |
== |
mx+3x+tmx+3x+t |
-3x+3x−3x+3x cancels out |
77 -t−t |
== |
mx+3x+tmx+3x+t -t−t |
Subtract tt from both sides |
7-t7−t |
== |
mx+3xmx+3x |
t-tt−t cancels out |
7-t7−t |
== |
mx+3xmx+3x |
7-t7−t |
== |
x(m+3)x(m+3) |
Finally, divide both sides by m+3m+3
7-t7−t |
== |
x(m+3)x(m+3) |
(7-t)(7−t)÷(m+3)÷(m+3) |
== |
x(m+3)x(m+3)÷(m+3)÷(m+3) |
|
7-tm+37−tm+3 |
== |
xx |
|
xx |
== |
7-tm+37−tm+3 |
Interchange the sides |
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Question 4 of 5
Write xx as the subject of the equation below
1-xx+y=11−xx+y=1
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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
Multiply both sides to x+yx+y
1-xx+y1−xx+y |
== |
11 |
|
1-xx+y1−xx+y×(x+y)×(x+y) |
== |
11×(x+y)×(x+y) |
|
1-x1−x |
== |
x+yx+y |
1x+y×(x+y)1x+y×(x+y) cancels out |
Leave only the xx terms on the right side
1-x1−x |
== |
x+yx+y |
1-x1−x +x+x |
== |
x+yx+y +x+x |
Add xx to both sides |
11 |
== |
2x+y2x+y |
-x+x−x+x cancels out |
11 -y−y |
== |
2x+y2x+y -y−y |
Subtract yy from both sides |
1-y1−y |
== |
2x2x |
y-yy−y cancels out |
Finally, divide both sides by 22
1-y1−y |
== |
2x2x |
(1-y)(1−y)÷2÷2 |
== |
2x2x÷2÷2 |
|
1-y21−y2 |
== |
xx |
2÷22÷2 cancels out |
|
xx |
== |
1-y21−y2 |
Interchange the sides |
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Question 5 of 5
Write bb as the subject of the equation below
x=2b3b+yx=2b3b+y
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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
Multiply both sides to 3b+y3b+y
xx |
== |
2b3b+y2b3b+y |
|
xx×(3b+y)×(3b+y) |
== |
2b3b+y2b3b+y×(3b+y)×(3b+y) |
|
x(3b+y)x(3b+y) |
== |
2b2b |
13b+y×(3b+y)13b+y×(3b+y) cancels out |
3bx+xy3bx+xy |
== |
2b2b |
Distribute xx |
Leave only bb terms on the left side
3bx+xy3bx+xy |
== |
2b2b |
3bx+xy3bx+xy -xy−xy |
== |
2b2b -xy−xy |
Subtract xyxy from both sides |
3bx3bx |
== |
2b-xy2b−xy |
xy-xyxy−xy cancels out |
3bx3bx -2b−2b |
== |
2b-xy2b−xy -2b−2b |
Subtract 2b2b from both sides |
3bx-2b3bx−2b |
== |
-xy−xy |
2b-2b2b−2b cancels out |
3bx-2b3bx−2b |
== |
-xy−xy |
b(3x-2)b(3x−2) |
== |
-xy−xy |
Finally, divide both sides by 3x-23x−2
b(3x-2)b(3x−2) |
== |
-xy−xy |
b(3x-2)b(3x−2)÷(3x-2)÷(3x−2) |
== |
-xy−xy÷(3x-2)÷(3x−2) |
|
bb |
== |
-xy3x-2−xy3x−2 |
(3x-2)÷(3x-2)(3x−2)÷(3x−2) cancels out |