# Bk5_Mock_6_E

Suppose u varies inversely as v. When v = 6, u = 3.

(a)Express u in terms of v.

(b)If u > 2, find the range of the values of v.

(c)If v is increased by 25%, find the percentage change in u.

For part c, new value of u = 18/1.25v = 1/1.25u = 0.8u

I can't understand why the new value of u is caluclated in this way???!!

Given that z varies directly as and y varies inversely as x.

Which of the following is true?

A.z varies inversely as x.

B.z varies directly as x.

C.z varies inversely as the square of x.

D.z varies directly as the square of x.

Which of the following may be the graph of y = 2.5-x + 2? (-x is the power of 2.5)

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(a)

Since u varies inversely as v,

then u=k/v (where k is a non-zero constant)

sub v = 6, u = 3 into u=k/v

3=k/6 --> k=18

so u=18/v

(b)

since u>2

then 18/v > 2 (v is a non-zero positive constant)

so 9>v>0

(c)

Since v is increased by 25%

New v = (1+25%)v = 1.25v

New u = 18 / (1.25)v = 14.4 / v

% change = {(14.4/v) - (18/v)}/ (18/v) x 100% = -20%

So u is decreased by 20%

For M.C.Q. 2

y = 2.5-x + 2

We can modify it to y=(2/5)^x + 2

It is an exponential function

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