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McGraw Hill Glencoe Algebra 2, 2012

MH

McGraw Hill Glencoe Algebra 2, 2012 View details

Standardized Test Practice

Exercise 1 Page 588

By dividing 1 by the time Greg or his father take to mow the lawn, we can find their speeds. What is the speed of Greg and his father working together?

C

Practice makes perfect

We are told that Greg's father can mow the lawn on his riding mower in 45 minutes. Dividing 1 lawn by 45 minutes, we can find the speed of Greg's father, or what part of the lawn he is mowing during one minute. Greg's Father: 1/45 lawn/min Similarly, we can calculate Greg's speed knowing that it takes him 1 hour 45 minutes, or 60+45=106 minutes to mow the lawn with a push mower. Greg: 1/105 lawn/min

Adding these speeds, we can find their speed of mowing when working together. Together: 1/45+1/105 lawn/min This expression tells us what part of the lawn Greg and his father can mow in one minute. How can we find the number of minutes t it would take them to mow the entire lawn working together? cc Lawn & & Time [0.3em] 145+ 1105 &-& 1 min [1em] 1 & - & t min Since the part of the lawn done varies directly as the time Greg and his father worked, we can set up the following proportion. 145+ 1105/1=1/t Dividing a number by 1 always results in the same number, so we can simplify the left-hand side of the equation by writing 145+ 1105 instead of the fraction. We will also multiply both sides of the equation by t to get rid of the fraction of the right-hand side. 145+& 1105=1/t [0.7em] & ⇓ [0.3em] t45+& t105=1 The answer is C.

Subchapter links

Exercises p.588-589