Changing The Subject

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#13887

Change the subject of the equation  to r: S = 500(1 – r).

$$\frac{S}{500} = 1 – r$$            divide both sides by 500

$$\frac{S}{500} +r = 1$$            add r to both sides

$$r = 1-\frac{S}{500}$$             subtract S over 500 from both sides

[Admin]

#13888

Given that $$\frac{b^2}{a^2}= 1-e^2$$ , a = 4 and e = 0.3, find the value of b correct to two decimal places.

$$\frac{b^2}{4^2}=1-0.3^2$$

$$\frac{b^2}{16}=0.91$$

$$\frac{b^2}{16}^{\times\cancel{16}}=0.91^{\times16}$$

b² = 14.56

b = √14.56

b = 3.82

[Admin]

#13889

Make v the subject of the formula: u = pq + vt².

u = pq + vt². u – pq = vt²             subtract pq from both sides

$$\frac{u-pq}{t^2}=v$$     divide both side by t²

$$v=\frac{u-pq}{t^2}$$

[Admin]

#18953

Change the subject of y = mx + b to x.
y = mx + b	we are trying to get x by itself, so we must remove all the other letters and terms. Start by subtracting b from both sides
y – b = mx	Divide both sides by m
##\frac{y-b}{m}=x##	swap sides so the x is on the left
##x=\frac{y-b}{m}##

[Admin]

#18954

The heat (Q) required to raise the temperature of a steel rod, of mass (m), from a temperature of A to a temperature of (B) is given by Q = 5m(B – A). Rearrange formula to make A the subject.
Q = 5m(B – A)	we are trying to get A by itself, so we must remove all the other letters and terms. Start by dividing both sides by 5m
##\frac{Q}{5m}=B-A##	Add A to both sides so that we can get rid of the negative sign
##\frac{Q}{5m}+A=B##	subtract $$\frac{Q}{5m}$$ from both sides to leave A by itself
##A= B-\frac{Q}{5m}##	we could combine the fraction with the lowest common denominator
##A=\frac{5mB-Q}{5m}##

[Admin]

#18956

Make s the subject of  v² = u² + 2as.
v² – u² = 2as	subtract u2 from both sides
##\frac{v^2-u^2}{2a}=\frac{\cancel{2a}s}{\cancel{2a}}##	divide both sides by 2a
##s=\frac{v^2-u^2}{2a}##

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