9,500 is 15% less than the original price.
So it must be 85% of the original price.
£9,500 = 0.85 P
Divide each side by 0.85 :
P = £9,500 / 0.85 = £11,176.47
Answer:
Step-by-step explanation:
Given are two lines equation in parametric form as
![l1: x=3-2t, y=7+4t, z=-3+8t\\l2: x=-1-u, y=18+3u, z=7+2u](https://tex.z-dn.net/?f=l1%3A%20x%3D3-2t%2C%20y%3D7%2B4t%2C%20z%3D-3%2B8t%5C%5Cl2%3A%20x%3D-1-u%2C%20y%3D18%2B3u%2C%20z%3D7%2B2u)
The direction ratios of the I line are -2, 4 and 8
Ii line are -1, 3, 2
These two are not proportional
Hence l1 and l2 are not parallel
Either they are skew or intersect
If intersect common point P would have coordinates as
![3-2t =- 1-u : 2t-u =4 \\7+4t = 18+3u: 4t-3u = 11\\-3+8t = 7+2u\\8t-2u = 10](https://tex.z-dn.net/?f=3-2t%20%3D-%201-u%20%20%3A%20%202t-u%20%3D4%20%5C%5C7%2B4t%20%3D%2018%2B3u%3A%20%20%204t-3u%20%3D%2011%5C%5C-3%2B8t%20%3D%207%2B2u%5C%5C8t-2u%20%3D%2010)
Let us solve first two equatins for u and t and check whether 3rd equation is satisfied by this.
2*(i) -ii gives u = -3
t =0.5
Substitute in III equation
8(0.5)-2(-3) = 10 so satisifed
These two are not skew lines
The point of intersection is got by substituting either t or u
Point of intersection (2, 9, 1)
Answer:
The graph is attached.
Step-by-step explanation:
The graph of the line with the slope
and the y-intercept
, has the form
![y=mx+b](https://tex.z-dn.net/?f=y%3Dmx%2Bb)
In our case the rate of change
is
; therefore we have the equation
![y=0.4x](https://tex.z-dn.net/?f=y%3D0.4x)
The graph of which is attached.
<em>P.S: we have set </em>
<em> by default, because we are not given any information for it. </em>
It will be b p/3r=q cause the others won't make sence
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 ![P_1U{(u,v)}UP_2](https://tex.z-dn.net/?f=P_1U%7B%28u%2Cv%29%7DUP_2)
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).