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McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

3. Simplifying Radical Expressions

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Exercise 51 Page 249

Start by solving the given formula for h.

Distance	3	6	9	12	15
Height	6	24	54	96	150

Practice makes perfect

Consider the given figure and formula.

The formula models the maximum distance d in miles that a lookout can see on a clear day from a height h. d=sqrt(3h/2)We are asked to determine the necessary height of the periscope if we want to see a ship from the submarine when the distance away from the ship is any of those listed in the table.

Distance	3	6	9	12	15
Height

Let's first solve the formula for h.

Solving the Formula

Since we have a square root, we will start by squaring both sides of the formula to solve it for h.

d=sqrt(3h/2)

LHS^2=RHS^2

d^2=(sqrt(3h/2))^2

( sqrt(a) )^2 = a

d^2=3h/2

LHS * 2=RHS* 2

2d^2=3h

.LHS /3.=.RHS /3.

2d^2/3=h

Rearrange equation

h=2d^2/3

With this formula, we can now find the corresponding heights of the given distances.

Finding the Height

To calculate the height for each distance, we need to substitute each value for d into the formula we found for h. Let's start with d= 3.

h=2d^2/3

d= 3

h=2( 3)^2/3

Calculate power

h=2(9)/3

Multiply

h=18/3

Calculate quotient

h=6

Proceeding in the same way, we can find the heights of the other distances.

Distance	2d^2/3	h
3	2( 3)^2/3	6
6	2( 6)^2/3	24
9	2( 9)^2/3	54
12	2( 12)^2/3	96
15	2( 15)^2/3	150

Finally, we can fill in the given table.

Distance	3	6	9	12	15
Height	6	24	54	96	150

Subchapter links

Exercises p.248-250