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suter [353]
3 years ago
15

Can someone please help me with number 37?

Mathematics
2 answers:
Gemiola [76]3 years ago
5 0

Answer: x=10

<u>Simplify both sides of the equation</u>

<u></u>4-(3x-5)=6-(2x+7)<u></u>

<u></u>4+-1(3x-5)=6+-1(2x+7) (Distribute the Negative Sign)

4+-1(3x)+(-1)(-5)=6+-1(2x)+(-1)(7)

4+-3x+5=6+-2x+-7

(-3x)+(4+5)=(-2x)+(6+-7) (Combine Like Terms)

-3x+9=-2x+-1

<u>Add 2x to both sides</u>

<u></u>-3x+9+2x=-2x-1+2x\\-x+9=-1<u></u>

<u></u>

<u>Subtract 9 from both sides</u>

<u></u>-x+9-9=-1-9\\-x=-10<u></u>

<u></u>

<u>Divide both sides by -1</u>

<u></u>-x/-1=-10/-1\\x=10<u></u>

Romashka [77]3 years ago
3 0

Answer:

<h2>x = 10</h2>

Step-by-step explanation:

4-(3x-5)=6-(2x+7)\\\\/-(a+b)=-1(a+b)=(-1)(a)+(-1)(b)-\text{distributive property}/\\\\4-3x-(-5)=6-2x-7\\\\4-3x+5=6-2x-7\qquad\text{combine like terms}\\\\-3x+(4+5)=-2x+(6-7)\\\\-3x+9=-2x-1\qquad\text{subtract 9 from both sides}\\\\-3x+9-9=-2x-1-9\\\\-3x=-2x-10\qquad\text{sdd}\ 2x\ \text{to both sides}\\\\-3x+2x=-2x+2x-10\\\\-x=-10\qquad\text{change the signs}\\\\\boxed{x=10}

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The Straight can start from 10 different positions: from an A, from a 2, 3, 4, 5, 6, 7, 8, 9 or from a 10 (if it starts from a 10, it ends in an A).

Given one starting position, we have 4 posibilities depending on the suit for each number, but we need to substract the 4 possible straights with the same suit. Hence, for each starting position there are 4⁵ - 4 possibilities. This means that we have 10 * (4⁵-4) = 10200 possibilities for a straight.

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We have 4 suits; each suit has 13 cards, so for each suit we have as many flushes as combinations of 5 cards from their group of 13. This is equivalent to the total number of ways to select 5 elements from a set of 13, in other words, the combinatorial number of 13 with 5 {13 \choose 5} .  However we need to remove any possible for a straight in a flush, thus, for each suit, we need to remove 10 possibilities (the 10 possible starting positions for a straight flush). Multiplying for the 4 suits this gives us

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We have 4 suits and 10 possible ways for each suit to start a straight flush. The suit and the starting position determines the straight flush (for example, the straight flush starting in 3 of hearts is 3 of hearts, 4 of hearts, 5 of hearts, 6 of hearts and 7 of hearts. This gives us 4*10 = 40 possibilities for a straight flush.

4 of a kind:

We can identify a 4 of a kind with the number/letter that is 4 times and the remaining card. We have 13 ways to pick the number/letter, and 52-4 = 48 possibilities for the remaining card. That gives us 48*13 = 624 possibilities for a 4 of a kind.

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We need to pick the pair of numbers that is repeated, so we are picking 2 numbers from 13 possible, in other words, {13 \choose 2} = 78 possibilities. For each number, we pick 2 suits, we have {4 \choose 2} = 6 possibilities to pick suits for each number. Last, we pick the remaining card, that can be anything but the 8 cards of those numbers. In short, we have 78*6*6*(52-8) = 123552 possibilities.  

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We choose the number that is matching from 13 possibilities, then we choose the 2 suits those numbers will have, from which we have 4 \choose 2 possibilities. Then we choose the 3 remaining numbers from the 12 that are left ( 12 \choose 3 = 220 ) , and for each of those numbers we pick 1 of the 4 suits available. As a result, we have

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We pick the suit that appears 3 times (4 possibilities), the one that appears twice (3 remaining possibilities). Foe the first suit we need 3 numbers from 13, and from the second one 2 numbers from 13 (It doesnt specify about matching here). This gives us

4 * 13 \choose 3 * 3 * 13 \choose 2 = 4*286*3*78 = 267696

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