The wording of your question suggests that there were answer choices. Mind sharing them?
The system <span>y=-2x^2 y=x-2 would be best written as:
</span><span>y=-2x^2
y=x-2 If you subst. x-2 for y in the first equation, you'll get:
x-2 = -2x^2, or 2x^2 + x - 2 = 0.
</span>
Answer:
-2+4i
Step-by-step explanation:
Simplify by combining the real and imaginary parts of each expression.
Okay so this friend has a 8, 9, 10 as her options that she pays at.
8: 2,6 9: 6,3 10: 5,5
6,2 3,6 5,5
4,4
4,4
These are the possibilities of rolling these numbers. = 4 possibilities
4/11 (11 because that is the number of possibilities you get for two dice)
that leaves 7/11 possibilities of rolling and not paying!
36% lands on a spot that she pays at and about 64% possibilities of not paying.
Hope this helped!
:)
Answer:
0, 3/4
Step-by-step explanation:
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
