Pearson Algebra 1 Common Core, 2011
PA
Pearson Algebra 1 Common Core, 2011 View details
3. Real Numbers and the Number Line
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Exercise 70 Page 22

Substitute the known values in the given formula and simplify.

4108.6 feet

Practice makes perfect
We are given a formula which can be used to calculate the length for a wire attached to an antenna at a height h. sqrt((h)^2 + (0.55h)^2) We want to find the value of this expression for the three different heights given. We can substitute these values of h into the given expression and simplify following the order of operations. Let's start with h= 200.
sqrt((h)^2 + (0.55h)^2)
sqrt(( 200)^2 + (0.55* 200)^2)
Simplify

Multiply

sqrt((200)^2 + (110)^2)
sqrt(40 000 + 12 100)
sqrt(52 100)
228.254244...

Now, remember that the problem says that 3 wires are used at each height, so we need to multiply our result by 3. (228.254244...) * 3 = 684.762732 ... Let's round the result to the nearest tenth. 684. 7 62732≈ 684.8 We can follow the same process to find the total length of wire needed for the other heights.

h sqrt((h)^2 + (0.55h)^2) Simplified * 3 Rounded to nearest tenth
200 sqrt(( 200)^2 + (0.55* 200)^2) 228.254244... 684.762732 ... 684.8
400 sqrt(( 400)^2 + (0.55* 400)^2) 456.508488... 1369.525465... 1369.5
600 sqrt(( 600)^2 + (0.55* 600)^2) 684.762732... 2054.288197... 2054.3

Finally, we need to add the lengths needed for each of the three heights to calculate the total amount of wire needed. 684.8 + 1369.5 + 2054.3 = 4108.6 Therefore, we need a minimum of 4108.6 feet of wire.