Question

A Number P Varies Directly As q And Partly Inversely As Q ^ 2, Given That P=11 When q=2 and p=25.16 when q=5. calculate the value of p when q=7

Answer 1

the description tells us that we have an equation of the form

p = mq + n/q^2

Now just plug in your values to find m and n.

11 = 2m+n/4
25.16 = 5m+n/25

That gives you a final equation of

p = 5q + 4/q^2

Now just plug in q=7 to find p

Answer 2

Since it varies directly as q and partly inversely as q^2 it is represented has this below
p = aq + b/q^2
The constants of proportionality here are a and b
11 = 2a + b/4 ----- i x 5
25.16 = 5a + b/25 ------ ii x 2
Multiply equation i by 5 and equation ii by 2
55 = 10a + 5b/4 ----- iii
50.32 = 10a + 2b/25 ----- iv
Subtract equation iv from equation iii
b = 4
Put the value of b in equation iv
50.32 = 10a + 8/25
Multiply through by 25
1258 = 250a + 8
a = 5
p =5q + 4/q^2
Therefore using this equation put into the equation the value of q which is 7
p = 5(7) + 4/(7)^2
p = 35 +4/49
p = 35.082

Answer 3

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