By the transversal theorem, if segments on a transversal cut by parallel lines are equal, then similar segments cut by the same parallel lines on another transversal are also equal.
Therefore
JK=KL=LM=5
=>
JM=3*5=15
Liza is driving to her sister's house 360 miles away. After 4 hours, Liza is 2/3 of the way there.
How many more hours does Liza have to drive?
<em><u>Answer:</u></em>
Liza need to drive 2 more hours to reach her sister house
<em><u>Solution:</u></em>
Given that Liza is driving to her sister's house 360 miles away
So total distance = 360 miles
After 4 hours, Liza is 2/3 of the way there.
![\frac{2}{3} r d \text { of total distance }=\frac{2}{3} \times 360=240](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B3%7D%20r%20d%20%5Ctext%20%7B%20of%20total%20distance%20%7D%3D%5Cfrac%7B2%7D%7B3%7D%20%5Ctimes%20360%3D240)
So Liza has covered 240 miles in 4 hours
Let us find the speed
<em><u>The relation between distance and speed is given as:</u></em>
![speed = \frac{distance}{time}](https://tex.z-dn.net/?f=speed%20%3D%20%5Cfrac%7Bdistance%7D%7Btime%7D)
![speed = \frac{240}{4} = 60](https://tex.z-dn.net/?f=speed%20%3D%20%5Cfrac%7B240%7D%7B4%7D%20%3D%2060)
Thus speed = 60 miles per hour
Assuming speed is same throughout the journey, let us calculate the time taken to complete remaining distance
Remaining distance = 360 - 240 = 120 miles
Now distance = 120 miles and speed = 60 miles per hour
![time taken = \frac{distance}{speed}\\\\time taken = \frac{120}{60} = 2](https://tex.z-dn.net/?f=time%20taken%20%3D%20%5Cfrac%7Bdistance%7D%7Bspeed%7D%5C%5C%5C%5Ctime%20taken%20%3D%20%5Cfrac%7B120%7D%7B60%7D%20%3D%202)
Thus Liza need to drive 2 more hours to reach her sister house
Answer:
The Depth of the lake had increased by 19%.
Step-by-step explanation:
Given:
Depth of lake few months ago = 1300 ft
depth of lake currently = 1547 ft
We need to find the percent of increase in depth of lake.
Solution:
First we will find the increase in depth of lake.
Increase in depth of lake can be calculated by subtracting Depth of lake few months ago from depth of lake currently.
framing in equation form we get;
increase in depth of lake = ![1547-1300= 247\ ft](https://tex.z-dn.net/?f=1547-1300%3D%20247%5C%20ft)
Now to find the percent of increase in depth of lake we will divide increase in depth of lake from Depth of lake few months ago and then multiply by 100.
framing in equation form we get;
percent of increase in depth of lake = ![\frac{247}{1300}\times 100 = 19\%](https://tex.z-dn.net/?f=%5Cfrac%7B247%7D%7B1300%7D%5Ctimes%20100%20%3D%20%2019%5C%25)
Hence the Depth of the lake had increased by 19%.
8.16 in
20% of 10.2 is 2.04
10.0-2.04 is 8.16 in