Which point satisfies both ƒ(x) = 2x and g(x) = 3x? (0,1) (0,-1) (1,0) (-1, 0)
2 answers:
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Answer:
Step-by-step explanation:
<h3>to understand this</h3><h3>you need to know about:</h3>- algebra
- PEMDAS
- A=2x+3
- B=x²-5x+1
- A-5B
A-5B=2x+3-5(x²-5x+1)
- distribute -5 to the parentheses:2x+3-5x²+25x-5
- rewrite:-5x²+2x+25x-5+3
- combine like terms:-5x²+27x-2
and
we are done
Answer:
The values of a, b and c are:
- a = 3
- b = 1
- c = -3
Step-by-step explanation:
Step 1:
First Confirm the sequence is quadratic by finding the second difference.
Sequence =
1st difference:
- 11-1=10
- 27-11=16
- 49-27=22
2nd difference:
- 16-10=6
- 22-16=6
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:
- a = 3
- b = 1
- c = -3