For this case we have the following table:
x f(x)
0 2
1 5
2 10
3 17
 The equation that best fits the data in the table, for this case, is given by a quadratic function.
 The quadratic function in its standard form is:
f (x) = x2 + 2x + 2
Answer:
f (x) = x2 + 2x + 2

8 0

3 years ago

Read 2 more answers

Ravi has 33 marbles his brother has twice as many how many marbles do they have all together

ycow [4]

Answer

Together, they have 99 marbles.

Explanation

Ravi has 33 marbles.

His brother has twice (two times) as many marbles as him. Ravi's brother has 33×2=66 marbles.

Together, they have 33+66=99 marbles.

6 0

3 years ago

On a ship the ratio of crewmembers to passengers os 1:5 there are 540 passengers how many crewmembers are there

goldenfox [79]

Answer: I believe that there are 108 crewmembers on the ship.

Step-by-step explanation: Since for every 5 passengers you have 1 crewmember i divided 540 by 5 landing me at the answer 108.

8 0

2 years ago

Enter the coefficients of the fifth Taylor polynomial T5(x) for the function f(x) = x5−3x4+2x2+5x−2 based at b=1. T5(x)= + (x−1)

DENIUS [597]

Compute the necessary values/derivatives of $f(x)$ at $x=1$ :

$f(1)=3$

$f'(1)=2$

$f''(1)=-12$

$f'''(1)=-12$

$f^{(4)}(1)=48$

$f^{(5)}(1)=120$

Taylor's theorem then says we can "approximate" (in quotes because the Taylor polynomial for a polynomial is another, exact polynomial) $f(x)$ at $x=1$ by

$T_5(x)=\dfrac3{0!}+\dfrac2{1!}(x-1)-\dfrac{12}{2!}(x-1)^2-\dfrac{12}{3!}(x-1)^3+\dfrac{48}{4!}(x-1)^4+\dfrac{120}{5!}(x-1)^5$

$T_5(x)=3+2(x-1)-6(x-1)^2-2(x-1)^3+2(x-1)^4+(x-1)^5$

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Another way of doing this would be to solve for the coefficients $a,b,c,d,e,g$ in

$f(x)=a+b(x-1)+c(x-1)^2+d(x-1)^3+e(x-1)^4+g(x-1)^5$

by expanding the right hand side and matching up terms with the same power of $x$ .

5 0

3 years ago