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user100 [1]
3 years ago
6

10n -11 =3 + 4n + 6n​

Mathematics
1 answer:
MrRissso [65]3 years ago
3 0

Answer:

10n -11 =3 + 4n + 6n​

10n - 10n -11 =3-3 + 4n + 6n​

-11 - 3 = -10n + 4n + 6n

-14= -10n + 10n

-14 = n

Step-by-step explanation:

10n -11 =3 + 4n + 6n​

10n - 10n -11 =3-3 + 4n + 6n​

-11 - 3 = -10n + 4n + 6n

-14= -10n + 10n

-14 = n

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82*

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disa [49]
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4 0
4 years ago
Given the set points K+1:(-1,8),(1,0) and (2,5), find the quadratic polynomial interpolate
Sauron [17]

Answer:

The interpolating polynomial is p(x) = 1-4x+3x^2.

Step-by-step explanation:

We want to find a quadratic polynomial p(x) such that p(-1)=8, p(1)=0 and p(2)=5. In order to do this let us write p(x) = a_0+a_1x+a_3x^2.

Now, evaluating the polynomial in the points -1, 1 and 2 we get

\begin{cases} 8 = p(-1) &= a_0-a_1+a_2\\ 0 = p(1) &= a_0+a_1+a_2\\ 5 = p(2) &= a_0+2a_1+4a_2\end{cases}

This relations give us a linear system of equations:

\begin{cases} 8 &= a_0-a_1+a_2\\ 0 &= a_0+a_1+a_2\\ 5&= a_0+2a_1+4a_2\end{cases}

where the a_0, a_1 and a_2 are the unknowns.

The augmented matrix of the system is

\begin{pmatrix}1 & -1 & 1 & 8\\ 1 & 1 & 1 & 0\\ 1 & 2 & 4 & 5\end{pmatrix}

In this matrix it is easy to eliminate the 1's of the first column and get

\begin{pmatrix} 1 & -1 & 1 & 8\\ 0 & 2 & 0 & -8\\ 0 & 3 & 3 & -3\end{pmatrix}

From this matrix we can find the values of each unknown. Notice that the second row gives us 2a_2=-8 that yields a_1=-4.

Then, the third row means 3a_1+3a_2=-3 that gives -12+3a_2=-3. So, a_2=3.

Finally, the first row is a_0-a_1+a_2=8 and substituting is a_0+7=8 that yields a_0=1.

Therefore, the interpolating polynomial is

p(x) = 1-4x+3x^2.

8 0
3 years ago
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