Answer the answer is f
Step-by-step explanation:
Ok so first, the skateboard costs $79.99. The skateboard costs 12% less with discount now so you have to pay 12% less.
You start by finding 12% of 79.99. This can be easily done by hand or with a calculator. Do 0.12 times 79.99. The answer is 9.5988. Since its 12% less we now subtract $9.5988 from $79.99. We get $70.3912. So this is the discounted price.
Now for the state sales tax. The sales tax is 6.75% which is equal to 0.0675. Like we did before, we multiplied .0675 by the value we wanted this percent of. But we can do a short way instead. Instead of getting 6.75% of $70.3912 and then adding it back to $70.3912, we can multiply $70.3912 by 1.0675 and this will do all the work for us.
So now, do $70.3912 times 1.0675 and you will have the answer. The answer is $75.142606, but wait, now we will round this to hundredths place as cents place ends their, and so the final answer will be $75.14.
Answer:
x = 46
Step-by-step explanation:
Since the angles are equal to each other, make the equations equal to each other.
4x - 36 = 3x + 10 and simplify
x - 36 = 10
x = 46
Hope this helps!
Answer:
Step-by-step explanation:
Given: MN ≅ MA
ME ≅ MR
Prove: ∠E ≅ ∠R
From the given diagram,
YN ≅ YA
EY ≅ RY
<EMA = <RMN (right angle property)
EA = EY + YA (addition property of a line)
NR = YN + RY (addition property of a line)
EA ≅ NR (congruent property)
ΔEMA ≅ ΔRMN (Side-Side-Side, SSS, congruence property)
<MNR ≅ MAE (angle property of congruent triangles)
Therefore,
<E ≅ <R (angle property of congruent triangles)
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).
