Given that
(2/x)+(3/y) = 13--------(1)
(5/x)-(4/y) = -2 -------(2)
Put 1/x = a and 1/y = b then
2a + 3b = 13 ----------(3)
On multiplying with 5 then
10a +15 b = 65 -------(4)
and
5a -4b= -2 ----------(5)
On multiplying with 2 then
10 a - 8b = -4 -------(6)
On Subtracting (6) from (4) then
10a + 15b = 65
10a - 8b = -4
(-)
_____________
0 + 23 b = 69
______________
⇛ 23b = 69
⇛ b = 69/23
⇛ b =3
On Substituting the value of b in (5)
5a -4b= -2
⇛ 5a -4(3) = -2
⇛ 5a -12 = -2
⇛ 5a = -2+12
⇛ 5a = 10
⇛ a = 10/5
⇛ a = 2
Now we have
a = 2
⇛1/x = 2
⇛ x = 1/2
and
b = 3
⇛1/y = 3
⇛ y = 1/3
<u>Answer :-</u>The solution for the given problem is (1/2,1/3)
<u>Check</u>: If x = 1/2 and y = 1/3 then
LHS = (2/x)+(3/y)
= 2/(1/2)+3/(1/3)
= (2×2)+(3×3)
= 4+9
= 13
= RHS
LHS=RHS is true
and
LHS=(5/x)-(4/y)
⇛ 5/(1/2)- 4/(1/3)
⇛(5×2)-(4×3)
⇛ 10-12
⇛ -2
⇛RHS
LHS = RHS is true
A combination is an unordered arrangement of r distinct objects in a set of n objects. To find the number of permutations, we use the following equation:
n!/((n-r)!r!)
In this case, there could be 0, 1, 2, 3, 4, or all 5 cards discarded. There is only one possible combination each for 0 or 5 cards being discarded (either none of them or all of them). We will be the above equation to find the number of combination s for 1, 2, 3, and 4 discarded cards.
5!/((5-1)!1!) = 5!/(4!*1!) = (5*4*3*2*1)/(4*3*2*1*1) = 5
5!/((5-2)!2!) = 5!/(3!2!) = (5*4*3*2*1)/(3*2*1*2*1) = 10
5!/((5-3)!3!) = 5!/(2!3!) = (5*4*3*2*1)/(2*1*3*2*1) = 10
5!/((5-4)!4!) = 5!/(1!4!) = (5*4*3*2*1)/(1*4*3*2*1) = 5
Notice that discarding 1 or discarding 4 have the same number of combinations, as do discarding 2 or 3. This is being they are inverses of each other. That is, if we discard 2 cards there will be 3 left, or if we discard 3 there will be 2 left.
Now we add together the combinations
1 + 5 + 10 + 10 + 5 + 1 = 32 choices combinations to discard.
The answer is 32.
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Note: There is also an equation for permutations which is:
n!/(n-r)!
Notice it is very similar to combinations. The only difference is that a permutation is an ORDERED arrangement while a combination is UNORDERED.
We used combinations rather than permutations because the order of the cards does not matter in this case. For example, we could discard the ace of spades followed by the jack of diamonds, or we could discard the jack or diamonds followed by the ace of spades. These two instances are the same combination of cards but a different permutation. We do not care about the order.
I hope this helps! If you have any questions, let me know :)
You do 65+90=155
then 180-155=25 to get the angle next to w
w=155
V(18-1811) también eres el mejor en la vida y tú me ganas de comer
If their equal(Same amount)
