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3 years ago

What is the result of dividing 2x3+x2−2x+8 by x + 2?

Mathematics

2 answers:

Irina-Kira [14]3 years ago

8 0

Answer:

Quotient = $2x^2 - 3x +4$

Remainder = 0

Step-by-step explanation:

Here, the given expression,

${2x^3 + x^2 - 2x + 8}\div {x + 2}$

By the long division, ( shown below )

We get,

$\frac{2x^3 + x^2 - 2x + 8}{x + 2}=(2x^2 - 3x + 4)\times (x+2) +0$

Since,

Dividend = Quotient × Divisor + Remainder

Hence,

Quotient = $2x^2 - 3x +4$

Remainder = 0

OLga [1]3 years ago

6 0

2x^3+x^2-2x+8 / x+2=

2x^2-3x+4

2x3+x2−2x+8 / x + 2=

6x+8 over x+2

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Read 2 more answers

5. Find the first, fourth, and tenth terms of the arithmetic

Lubov Fominskaja [6]

The first, fourth and tenth terms of the arithmetic sequence is $-6, -\frac{27}{5}$ and $-\frac{21}{5}$

Explanation:

The given rule for the arithmetic sequence is $A(n)=-6+(n-1)(\frac{1}{5} )$

We need to determine the first, fourth and tenth terms of the sequence.

To find the first, fourth and tenth terms, let us substitute $n=1,4,10$ in the general rule for the arithmetic sequence.

To find the first term, substitute $n=1$ in $A(n)=-6+(n-1)(\frac{1}{5} )$ , we get,

$A(1)=-6+(1-1)(\frac{1}{5} )$

$A(1)=-6+(0)(\frac{1}{5} )$

$A(1)=-6$

Thus, the first term of the arithmetic sequence is -6.

To find the fourth term, substitute $n=4$ in $A(n)=-6+(n-1)(\frac{1}{5} )$ , we get,

$A(2)=-6+(4-1)(\frac{1}{5} )$

$A(2)=-6+(3)(\frac{1}{5} )$

$A(2)=\frac{-30+3}{5}$

$A(2)=\frac{-27}{5}$

Thus, the fourth term of the arithmetic sequence is $-\frac{27}{5}$

To find the tenth term, substitute $n=10$ in $A(n)=-6+(n-1)(\frac{1}{5} )$ , we get,

$A(10)=-6+(10-1)(\frac{1}{5} )$

$A(10)=-6+(9)(\frac{1}{5} )$

$A(10)=-6+\frac{9}{5}$

$A(10)=-\frac{21}{5}$

Thus, the tenth term of the arithmetic sequence is $-\frac{21}{5}$

Hence, the first, fourth and tenth terms of the arithmetic sequence is $-6, -\frac{27}{5}$ and $-\frac{21}{5}$

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4 years ago

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Answer:

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The slope intercept form is:

mx + b = y

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The y intercept is 3.

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2 years ago

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Answer:

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