Find all the polynomials f (t ) of degree ≤ 2 [of the form f (t) = a + bt + ct2] whose graphs run through the points (1, 3) and
(2, 6), such that f ′(1) = 1 [where f ′(t) denotes the derivative?
1 answer:
Answer:
f(t) = 4 - 3t + 2t^2
Step-by-step explanation:
Given:
f(t) = a + bt + ct^2
Differentiate w.r.t. "t"
f'(t) = b + 2ct
1) f(1) = 3
a + b +c = 3...........1)
2) f(2) = 6
a + 2b + 4c = 6.........2)
3) f'(1) = 1
b + 2c = 1 ...............3)
Solving above three simultaneous equations:
a = 4
b = -3
c = 2
Therefore equation becomes:
f(t) = 4 + (-3)t + (2)t^2
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Step-by-step explanation:
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Answer:
The length is 6, and the width is 4.
Step-by-step explanation:
The length and width add up to 10. 6 + 4 = 10.
A quarter of the width plus the length = 7. A quarter of 4 = 1. 1 + 6 = 7
Las razones de los lados de un triángulo rectángulo se llaman razones trigonométricas.
Espero te ayude :)
Answer:
28 craft sticks
Step-by-step explanation:
Do 84/45
You get 1.8666
For every 1 pom pom you have 1.8666 craft sticks
Now you can multipy 15 by 1.8666 to get 28!