To factor quadratic equations of the form ax^2+bx+c=y, you must find two values, j and k, which satisfy two conditions.
jk=ac and j+k=b
The you replace the single linear term bx with jx and kx. Finally then you factor the first pair of terms and the second pair of terms. In this problem...
2k^2-5k-18=0
2k^2+4k-9k-18=0
2k(k+2)-9(k+2)=0
(2k-9)(k+2)=0
so k=-2 and 9/2
k=(-2, 4.5)
8. It is a function, the x’s do not repeat
9. It is not a function, -3 is listed twice
Answer:
12 home runs and brainlypatrol don't delete this question!!!!!
Step-by-step explanation:
<h3>
Answer: 82</h3>
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Explanation:
Let A be the tens digit and B be the units digit
We can write the number as AB. So if A = 9 and B = 2 for instance, then the number is AB = 92. Keep in mind I'm not multiplying A and B here.
"The unit digit is a prime" means B could be any of these values {2,3,5,7}. We only list the single digit primes. The value 1 is not prime.
Now multiply each of those items by 4
4*2 = 8
4*3 = 12
4*5 = 20
4*7 = 28
Of those results, only 4*2 = 8 leads to a single digit answer.
This must mean that A = 8 and B = 2
Therefore the number is AB = 82.
