<span>Im pretty sure that it's 0
</span>
Your answer is Graph A.
Hope this helps :)!!!
Answer:
(x, y) → (x + 6, y + 13)
Step-by-step explanation:
<u>Let's use one of the points to determine the transition:</u>
- Coordinates of D = (-5, -7)
- Coordinates of D' = (1, 6)
<u>Transition:</u>
- x' - x = 1 - (-5) = 6
- y' - y = 6 - (-7) = 13
<u>So the rule is:</u>
To solve using synthetic division:
Draw an upside down division symbol (I did the best I can given the format). Place the constant of x - 1 (1) on the outside, and the coefficients of x³ - 2x² - 1 on the inside, like so:
| x³ x² x¹ x⁰
1 | 1 -2 0 -1
|
|___________
Next, bring down the first coefficient.
| x³ x² x¹ x⁰
1| 1 -2 0 -1
|
|___________
1
Multiply the number on the outside of the symbol (1) by the coefficient you've just brought down (1). Place that number under the next coefficient to the right.
| x³ x² x¹ x⁰
1| 1 -2 0 -1
| 1
|___________
1
Add the new number (1) to the coefficient above it (-2) and place that number directly underneath it.
| x³ x² x¹ x⁰
1| 1 -2 0 -1
| 1
|___________
1 -1
Continue this process until you've run out of numbers to multiply.
| x³ x² x¹ x⁰
1| 1 -2 0 -1
| 1 -1 -1
|___________
1 -1 -1 -2
The resulting numbers are coefficients. The first three are part of the quotient, and the last one is part of the remainder. The remainder becomes the last coefficient (-2) over the divisor (x - 1).
Answer:
(x³ - 2x² - 1) / (x - 1) = x² - x - 1 + -2/x - 1
(10g^3 - 8g^2 + g - 14)(10g^2 + g)
Multiply eah term by 10g^2 and then add it to each term times g
100g^5 - 80g^4 + 10g^3 - 140g^2 + 10g^4 - 8g^3 + g^2 - 14g)
Combine like terms
100g ^5 - 80g^4 + 10g^4 + 10g^3 - 8g^3 - 140g^2 + g^2 - 14g
100g^5 - 70g^4 + 2g^3 - 139g^2 - 14g
Answer: 100g^5 - 70g^4 + 2g^3 - 139g^2 - 14g
hope it helps :)
