$122.57 • 4 + $555.28 = $1,045.56
$136.71 • 4 + $585.34 = $1,132.18
$1,132.18 - $1,045.56 = $86.62
Answer:
a) 2.5
b) 6.25
Step-by-step explanation:
For similar figures, the ratio of any corresponding linear dimensions is the same. The ratio of areas is the square of that.
<h3>Application</h3>
The ratio of linear dimensions, larger to smaller, is ...
(30 yd)/(12 yd) = 2.5
<h3>a) Perimeter</h3>
Perimeter is a linear dimension, the sum of side lengths. The ratio of perimeters is 2.5.
<h3>b) Area</h3>
The ratio of areas, larger to smaller, is the square of the scale factor for side lengths:
(2.5)² = 6.25
The ratio of the areas of the larger to smaller figure is 6.25.
2,3,7,and 5 are the factors
To find the decimal form, you have to manually divide 1 by 8. Since 8 is greater than 1, the quotient would then start with 0., then you add a 0 next to 1, to make it 10. This time, divide 10 by 8. The nearest answer would be 1, because 1 *8 = 8. Subtracting this from 10, you get 2. Add another 0 to 2, to make it 20. Do the cycle all over again. The complete solution is as follows:
0.125
---------------------------
8 | 10
- 8
------------------------
20
-16
-------------
40
- 40
-------------
0
<em>Hence, the decimal form of 1/8 is 0.125.</em>
It's not obvious here, but you're being asked to find a linear equation for the velocity of the car, given two points on the line that represents this velocity.
Find the slope of the line segment that connects the points (3 hr, 51 km/hr) and (5 hr, 59 km/hr). Graph this line. Where does this line intersect the y-axis? Find the y-value; it's your "y-intercept," b.
Now write the equation: velocity = (slope of line)*t + b
The units of measurement of "slope of line" must be "km per hour squared," and those of the "y-intercept" must be "km per hour."
Part B: Start with the y-intercept (calculated above). Plot it on the vertical axis of your graph. Now label the horizontal axis in hours: {0, 1, 2, 3, 4, 5, 6}. Draw a vertical line through t=6 hours. It will intercept both the horiz. axis and the sloping line representing the velocity as a function of time. Show only the part of the graph that extends from t=0 hours to t=6 hours.