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3 years ago

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Mr. Jackson lengthened the fence along the back of his yard. Before he added 130 feet, the fence was 300 feet long. He divided t

he fence into 5-foot sections.
How many 5-foot sections does he have now?
A. 86
B. 60
C. 26
D. 430
is it A?

1 answer:

hram777 [196]3 years ago

8 0

Answer:

The answer is A, 86 sections.

Step-by-step explanation:

Step one : Add the 130 feet on to the 300 feet to find the new length of the fence.

130 + 300 = 430

Step two : find how many 5 foot sections he has now, divide 430 by 5.

430/5 = 86

Mr. Jackson can divide the new fence into 86 5-foot sections.

The answer is A, 86 sections.

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Answer:

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Step-by-step explanation:

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2. Plug in 0 for y in the equation.

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3 years ago

Given the set points K+1:(-1,8),(1,0) and (2,5), find the quadratic polynomial interpolate

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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

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

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Therefore, the interpolating polynomial is

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Answer:

<h2> $x = 3$ </h2>

Step-by-step explanation:

<h3> $4x − 5 = −17$ </h3>

<u>Add 5 to both sides of the equation</u>

That's

<h3> $4x + 5 - 5 = - 17 + 5 \\ 4x = - 12$ </h3>

<u>Divide both sides by 4</u>

<h3> $\frac{4x}{4} = - \frac{12}{4} \\$ </h3>

We have the final answer as

<h3> $x = 3 \\$ </h3>

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