Answer:
The values of a, b and c are:
Step-by-step explanation:
Step 1:
First Confirm the sequence is quadratic by finding the second difference.
Sequence = 
1st difference:
2nd difference:
Step 2:
Just divide the second difference by 2, you will get the value of 
.
6 ÷ 2 = 3
So the first term of the nth term is 
Step 3: Next, substitute the number 1 to 5 into
n = 1,2,3,4,5
3n² = 3,12,27,48,75
Step 4:
Now, take these values (3n²) from the numbers in the original number sequence and work out the nth term of these numbers that form a linear sequence.
n = 1,2,3,4,5
3n² = 3,12,27,48,75
Differences:
1 - 3 = -2
11 - 12 = -1
27 - 27 = 0
49 - 48 = 1
Now the nth term of these differences (-2,-1,0,1) is (n - 1) - 2.
so b = 1, and c = -3
Step 5: Write down your final answer in the form an² + bn + c.
3n² + (n - 1) -2
= 3n² + n - 1 - 2
= 3n² + n - 3
Therefore, the values of a, b and c are:
See solution in photo! Please mark me as brainliest and give a thanks if it helped :) have a nice day
X-intercepts are found by factoring. Use the quadratic formula on the first since it's in standard form and you find that your x values are in fact -3 and 7. For the second one, use the Zero Product Property that says that x - 3 = 0 or x + 7 = 0. Therefore, x = 3 and -7. Signs are wrong. So not the second one. As for the third one, if you factor out a 3, your polynomial is exactly the same as the first one which did give us the desired x values. So the third one checks out. If you FOIL out the first one and then apply the quadratic formula you do get x = 3 and -7. So the fourth one checks out too. For the last one, putting it into the quadratic formula gives you x values of 3 and -7, so no to that one. Summary: 1st, 3rd, 4th have zeros of -3 and 7; 2nd and 5th do not.
Answer:
oh ur up at night.. me to...
Step-by-step explanation:
Xy= 120
x= 7+y
(7+y)(y) = 120
y^2 + 7y - 120 = 0
(y + 15) (y - 8) = 0
Since we are finding two positive numbers,
If y= 8 then x= 15
The two numbers are 8 and 15
