Answer:
D. (27x^6y^9) / x^9.
Step-by-step explanation:
(3x^2y^3 / z^3 )^3
= 3^3 x^(2*3)y(3*3) / z^(3^3)
= 27 x^6y^9 / x^9.
Answer:
(2, 5).
Step-by-step explanation:
The length of the line joining the first 2 points = (5 - 3) = 2 units. This is a vertical line because the 2 x coordinates ( -6 and - 6) are equal.
The line joining ( -6 , 3) and (2 , 3) is horizontal and is 2 - -6 = 8 units long.
Because we have a rectangle the point we need to find must have x coordinate of 2 and the y coordinate will be 2 more than the y coordinate 3.
.So it is (2, 5),
The slope of this line is -1 you find this by using the slip formula of y-y over x-x find two points on the line so I used points (-2,2) and (-1,1) plug them into the formula to get the slope of -1 hope this helps:)
Answer:
a. p1(x) = 2 - x
b. p2(x) = x² - 3*x + 3
c. p1(0.97) = 1.03; p2(0.97) = 1.0309
Step-by-step explanation:
f(x) = 1/x
f'(x) = -1/x²
f''(x) = 2/x³
a = 1
a. The linear approximating polynomial is:
p1(x) = f(a) + f'(a)*(x - a)
p1(x) = 1/1 + -1/1² * (x - 1)
p1(x) = 1 - x + 1
p1(x) = 2 - x
b. The quadratic approximating polynomial is:
p2(x) = p1(x) + 1/2 * f''(a)*(x - a)²
p2(x) = 2 - x + 1/2 * 2/1³ * (x - 1)²
p2(x) = 2 - x + (x - 1)²
p2(x) = 2 - x + x² - 2*x + 1
p2(x) = x² - 3*x + 3
c. approximate 1/0.97 using p1(x)
p1(0.97) = 2 - 0.97 = 1.03
approximate 1/0.97 using p2(x)
p2(0.97) = 0.97² - 3*0.97 + 3 = 1.0309
Answer: 3p² + q² - 5pq - 10 < 0
<u>Step-by-step explanation:</u>
p² - q² + 2pq < -2p² - 2q² + 7pq + 10
since you didn't specify what to solve for, move everything from the right side to the left side of the inequality symbol using inverse operations.
(p² + 2p²) + (- q² + 2q²) + (2pq - 7pq) + (- 10) < 0
3p² + q² + -5pq + -10 < 0
3p² + q² - 5pq - 10 < 0
