Big Ideas Math Integrated I, 2016
BI
Big Ideas Math Integrated I, 2016 View details
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Exercise 27 Page 549

Practice makes perfect
a

We want to find out how many units the amoeba moves both horizontally and vertically between B3 and G7. First we move down from 3 to 7 and then to the right from B to G.

When the amoeba travels from B3 to G7, it moves 4 units down and 5 units to the right.

b

Now we want to find the distance d that the amoeba moved.

Since we know the horizontal and vertical length the amoeba moved, we can use the Distance Formula to calculate the distance. d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2) In this case, we already know the distance traveled in both directions so we can substitute these values into the places for the change in x and change in y.

d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
d=sqrt(5^2+ 4^2)
d=sqrt(25+16)
d=sqrt(41)
d≈ 6.4

Finally, remember that each grid unit represents 2 millimeters so we need to multiply this distance by 2. 6.4* 2 = 12.8 The amoeba travels about 12.8 millimeters.

c

In Part B, we calculated that the amoeba traveled about 12.8 millimeters. To calculate the speed in millimeters per second, we should divide the distance by the time. The time it took for the amoeba from B3 to G7 was 24.5. Therefore, the speed can be calculated as follows.

12.824.5=0.52244...≈0.52 The amoeba moves at a speed of 0.52 millimeters per second.