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Harrizon [31]
3 years ago
9

For which equation does it make the most sense to solve by using the zero product property?

Mathematics
2 answers:
Svetach [21]3 years ago
6 0

Answer:

Option 1 - x^2+11x+30=0

Step-by-step explanation:

We have to find : For which equation does it make the most sense to solve by using the zero product property?

Equations :

1) x^2+11x+30=0

2) x^2+2x=119

3) 4x^2+-2x=15

4) 2x^2+2x-9=0

In the following equation only equation 1 is factories easily but rest 2,3,4 equations are prime equation they solve by quadratic formula but not by zero product property,

As In equation 1,

x^2+11x+30=0

(x+5)(x+6)=0

Applying zero product property,

i.e, If ab=0 then either a=0 or b=0

So, Either (x+5)=0 or (x+6)=0

Either x=-5 or x=-6

Therefore, Option 1 is correct.

mafiozo [28]3 years ago
3 0
The 1st equation can be factored into
(x + 5)(x + 6) = 0
you can apply the zero product rule now.
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Answer:

a) 182 possible ways.

b) 5148 possible ways.

c) 1378 possible ways.

d) 2899 possible ways.

Step-by-step explanation:

The order in which the cards are chosen is not important, which means that we use the combinations formula to solve this question.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

In this question, we have that:

There are 52 total cards, of which:

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13 are diamonds.

13 are hearts.

13 are clubs.

(a)Two-pairs: Two pairs plus another card of a different value, for example:

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1 other card, from a set of 26(whichever two cards were not chosen above). So

T = 2C_{13,2} + C_{26,1} = 2*\frac{13!}{2!11!} + \frac{26!}{1!25!} = 182

So 182 possible ways.

(b)Flush: five cards of the same suit but different values, for example:

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(c)Full house: A three of a kind and a pair, for example:

4 combinations of 3 from a set of 13(three of a kind ,c an be all possible kinds).

3 combinations of 2 from a set of 13(the pair, cant be the kind chosen for the trio, so 3 combinations). So

T = 4*C_{13,3} + 3*C_{13.2} = 4*\frac{13!}{3!10!} + 3*\frac{13!}{2!11!} = 1378

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4 combinations of 4 from a set of 13(four of a kind, can be all spades, all diamonds, and hearts or all clubs).

1 from the remaining 39(do not involve the kind chosen above). So

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So 2899 possible ways.

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hello :<span>
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y = 14  or y = -16
you have two  points : (-15,14)  , (-15, -16)</span>
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