Answer:
it will be 36 (B)
Step-by-step explanation:
1/2(72+55)=36
hope this help!
Let the slower train's velocity be x-21
Let the faster train's velocity be x
We know that the approach speed is the sum of both speeds, so x+x -21= 2x-21.
The approach rate is given by Distance/time = 471/3 = 157mpH
x+x-21=157
2x=157+21
2x=178
x=89mph
The slower train is travelling 89-21 = 68mph
The faster train is travelling 89mph.
Answer:
x=10
y=-3
Step-by-step explanation:
3x + 5y= 15...eqn. 1
2x + 4y= 8...eqn. 2
multiply eqn. 1 by 2 and eqn. 2 by 3
6x + 10y = 30...eqn. 3
<u>6x</u><u> </u><u>+</u><u> </u><u>12y</u><u> </u><u>=</u><u> </u><u>24...eqn. 4</u>
0 - 2y = 6
divide both sides by the coefficient of y which is -2
<u>- 2y</u> = <u>6</u>
- 2 -2
y= -3
Put y= -3 into eqn. 1
3x + 5(-3) = 15
3x - 15 = 15
3x =15 + 15
<u>3x</u> =<u> 30</u>
3 3
x = 10
Therefore X= 10 and Y= -3
Answer: hello your question lacks some data hence I will be making an assumption to help resolve the problem within the scope of the question
answer:
≈ 95 units ( output level )
Step-by-step explanation:
Given data :
P = 2000 - Q/10
TC = 2Q^2 + 10Q + 200 ( assumed value )
<u>The output level where a purely monopolistic market will maximize profit</u>
<u>at MR = MC </u>
P = 2000 - Q/10 ------ ( 1 )
PQ = 2000Q - Q^2 / 10 ( aka TR )
MR = d (TR ) / dQ = 2000 - 2Q/10 = 2000 - Q/5
TC = 2Q^2 + 10Q + 200 ---- ( 2 )
MC = d (TC) / dQ = 4Q + 10
equating MR = MC
2000 - Q/5 = 4Q + 10
2000 - 10 = 4Q + Q/5
1990 = 20Q + Q
∴ Q = 1990 / 21 = 94.76 ≈ 95 units ( output level )
