We are able to set 4x+32 equal to 6x-8 because these are congruent vertical angles
4x+32=6x-8
4x+40=6x (add 8 to both sides)
40=2x (subtract 4x from both sides)
X=20 (divide both sides by 2)
Then we plug in 20 for x into 6x-8 to find the answer
6(20)-8
=120-8
=112
Hope this helps!
Answer:
<em>Rate of Pratap in still water is 4.5 miles/hour and rate of current is 0.5 miles/hour.</em>
Step-by-step explanation:
Pratap Puri rowed 10 miles down a river in 2 hours, but the return trip took him 2.5 hours.
We know that, 
So, the <u>speed of Pratap with the current</u> will be: 
miles/hour
and the <u>speed of Pratap against the current</u> will be: 
miles/hour.
Suppose, the rate of Pratap in still water is 
and the rate of current is 
.
So, the equations will be........
Adding equation (1) and (2) , we will get......
Now, plugging this 
into equation (1), we will get.....
Thus, Pratap can row at 4.5 miles per hour in still water and the rate of the current is 0.5 miles/hour.
For this case we have that the original value of the car is:
m dollars
For the following year we have the value is:
((100-15) / (100)) m
Rewriting we have:
((85) / (100)) m
0.85m
Answer:
the value of his car the year after the car is worth m dollars is:
B.f (m) = 0.85m
Answer:
Image result for 1 What are the names of four coplanar points?
1. What are the names of four coplanar points? A. B. C. D. Points P, M, F, and C are coplanar Points F, D, P, and N are coplanar.
Step-by-step explanation:
Problem 7)
The answer is choice B. Only graph 2 contains an Euler circuit.
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To have a Euler circuit, each vertex must have an even number of paths connecting to it. This does not happen with graph 1 since vertex A and vertex D have an odd number of vertices (3 each). The odd vertex count makes it impossible to travel back to the starting point, while making sure to only use each edge one time only.
With graph 2, each vertex has exactly two edges attached to it. So an Euler circuit is possible here.
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Problem 8)
The answer is choice B) 5
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Work Shown:
abc base 2 = (a*2^2 + b*2^1 + c*2^0) base 10
101 base 2 = (1*2^2 + 0*2^1 + 1*2^0) base 10
101 base 2 = (1*4 + 0*2 + 1*1) base 10
101 base 2 = (4 + 0 + 1) base 10
101 base 2 = 5 base 10
