Exercise 49 Page 249

Practice makes perfect

We are given that the amount of current in amperes I that an appliance uses can be calculated by the following formula.

I=sqrt(P/R) In this formula, P is the power in watts and R is the resistance in ohms. To simplify this formula, we will rationalize the denominator using the Quotient Property and the Product Property of Square Roots.

I=sqrt(P/R)

SqrtQuot

sqrt(a/b)=sqrt(a)/sqrt(b)

I=sqrt(P)/sqrt(R)

ExpandFrac

a/b=a * sqrt(R)/b * sqrt(R)

I=sqrt(P)* sqrt(R)/sqrt(R)* sqrt(R)

ProdSqrt

sqrt(a)*sqrt(b)=sqrt(a* b)

I=sqrt(PR)/sqrt(R* R)

ProdToPowTwoFac

a* a=a^2

I=sqrt(PR)/sqrt(R^2)

SqrtPowToNumber

sqrt(a^2)=a

I=sqrt(PR)/R

We are asked to calculate how much current an appliance uses if P is 75 watts and R is 5 ohms. To do so, let's substitute these values into the simplified form of the formula.

I=sqrt(PR)/R

SubstituteII

P= 75, R= 5

I=sqrt(75( 5))/5

Multiply

I=sqrt(375)/5

UseCalc

Use a calculator

I=3.872983...

RoundDec

Round to 1 decimal place(s)

I≈ 3.9