Answer:

Step-by-step explanation:
Hello,
a and b are the zeros, we can say that

So we can say that

Now, we are looking for a polynomial where zeros are 2a+3b and 3a+2b
for instance we can write

and we can notice that
so
![(x-2a-3b)(x-3a-2b)=x^2-5(a+b)x+6[(a+b)2-2ab]+13ab\\= x^2-5(a+b)x+6(a+b)^2+ab](https://tex.z-dn.net/?f=%28x-2a-3b%29%28x-3a-2b%29%3Dx%5E2-5%28a%2Bb%29x%2B6%5B%28a%2Bb%292-2ab%5D%2B13ab%5C%5C%3D%20x%5E2-5%28a%2Bb%29x%2B6%28a%2Bb%29%5E2%2Bab)
it comes

multiply by 3

Answer:
1/3, 5/12, 1/2, 3/4
Step-by-step explanation:
Answer:13
Step-by-step explanation: I took the test in K12 and the answer is 13
Answer:
nsdkbljcb;awjef
Step-by-step explanation:
jwaefjalfjwkhjefkuw
Answer:
Yes.
Step-by-step explanation:
Yes because each value of x maps to a single value of y.
THere is a many-to-one relation with (1, 12) and (5, 12) but that is OK for a function. What would not be OK is if we had a one-to-many relation like (1, 12) and (1, 14) included in the ordered pairs - that is not a function.