The line that passes through the point (4,9) and is parallel to y=3/4x+1
1 answer:
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Answer:
Check the explanation
Step-by-step explanation:
1) Algorithm for finding the new optimal flux: 1. Let E' be the edges eh E for which f(e)>O, and let G = (V,E). Find in Gi a path Pi from s to u and a path , from v to t.
2) [Special case: If , and
have some edge e in common, then Piu[(u,v)}uPx has a directed cycle containing (u,v). In this instance, the flow along this cycle can be reduced by a single unit without any need to change the size of the overall flow. Return the resulting flow.]
3) Reduce flow by one unit along
4) Run Ford-Fulkerson with this sterling flow.
Justification and running time: Say the original flow has see F. Lees ignore the special case (4 After step (3) Of the elgorithuk we have a legal flaw that satisfies the new capacity constraint and has see F-1. Step (4). FOrd-Fueerson, then gives us the optimal flow under the new cePacie co mint. However. we know this flow is at most F, end thus Ford-Fulkerson runs for just one iteration. Since each of the steps is linear, the total running time is linear, that is, O(lVl + lEl).
Answer:
Step-by-step explanation:
The standard form of a quadratic is
We will use the x and y values from each of our 3 points to find a, b, and c. Filling in the x and y values from each point:
First point (-5, 0):
and
0 = 25a - 5b + c
Second point (9, 0):
and
0 = 81a + 9b + c
Third point (8, -39):
and
-39 = 64a + 8b + c
Use the elimination method of solving systems on the first 2 equations to eliminate the c. Multiply the first equation by -1 to get:
-25a + 5b - c = 0
81a + 9b + c = 0
When the c's cancel out you're left with
56a + 14b = 0
Now use the second and third equations and elimination to get rid of the c's. Multiply the second equation by -1 to get:
-81a - 9b - c = 0
64a + 8b + c = -39
When the c's cancel out you're left with
-17a - 1b = -39
Between those 2 bolded equations, eliminate the b's. Do this by multiplying the second of the 2 by 14 to get:
56a + 14b = 0
-238a - 14b = -546
When the b's cancel out you're left with
-182a = -546 and
a = 3
Use this value of a to back substitute to find b:
56a + 14b = 0 so 56(3) + 14b = 0 gives you
168 + 14b = 0 and 14b = -168 so
b = -12
Now back sub in a and b to find c:
0 = 25a - 5b + c gives you
0 = 75+ 60 + c so
0 = 135 + c and
c = -135
Put that all together into the standard form equation to get