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McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

5. Radical Equations

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Exercise 1 Page 261

Substitute the given value into the formula for the surface area of a sphere.

r=sqrt(π x)/2π

Practice makes perfect

We are given a formula for the surface area of a sphere. SA=4π r^2

There is a basketball with a surface area of x square inches.

Sphere

We will find the radius r of this basketball by substituting SA= x into the formula.

SA=4π r^2

SV= x

x=4π r^2

Rearrange equation

4π r^2=x

▼

Solve for r

.LHS /4π.=.RHS /4π.

r^2=x/4π

sqrt(LHS)=sqrt(RHS)

r=sqrt(x/4π)

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

r=sqrt(x)/sqrt(4π)

a/b=a * sqrt(π)/b * sqrt(π)

r=sqrt(x)* sqrt(π)/sqrt(4π)* sqrt(π)

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

r=sqrt(xπ)/sqrt(4π^2)

Calculate root

r=sqrt(xπ)/2π

CommutativePropMult

Commutative Property of Multiplication

r=sqrt(π x)/2π

The radius of the basketball is sqrt(π x)2π inches.

Subchapter links

Exercises p.261-263