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3 years ago

Divide f(x) by d(x), and write a summary statement in the form indicated.

Mathematics

1 answer:

Mandarinka [93]3 years ago

3 0

Answer:

The correct option is B) $f(x) =(x^2+1)(x^2+4x+5)$

Step-by-step explanation:

Consider the provided function.

$f(x) = x^4 + 4x^3 + 6x^2 + 4x + 5$ and $d(x) = x^2+1$

We need to divide f(x) by d(x)

As we know: Dividend = Divisor × Quotient + Remainder

In the above function f(x) is dividend and divisor is d(x)

Divide the leading term of the dividend by the leading term of the divisor: $\frac{x^4}{x^2}=x^2$

Write the calculated result in upper part of the table.

Multiply it by the divisor: $x^2(x^2+1)=x^4+x^2$

Now Subtract the dividend from the obtained result:

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

Again divide the leading term of the obtained remainder by the leading term of the divisor: $\frac{4x^3}{x^2}=4x$

Write the calculated result in upper part of the table.

Multiply it by the divisor: $4x(x^2+1)=4x^3+4x$

Subtract the dividend:

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

Divide the leading term of the obtained remainder by the leading term of the divisor: $\frac{5x^2}{x^2}=5$

Multiply it by the divisor: $5(x^2+1)=5x^2+5$

Subtract the dividend:

$(5x^2+5)-(5x^2+5)=0$

Therefore,

Dividend = $x^4 + 4x^3 + 6x^2 + 4x + 5$

Divisor = $x^2+1$

Quotient = $x^2+4x+5$

Remainder = 0

Dividend = Divisor × Quotient + Remainder

$f(x) = (x^2+1)(x^2+4x+5)$

Hence, the correct option is B) $f(x) =(x^2+1)(x^2+4x+5)$

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Answer:

Hence, none of the options presented are valid. The plane is represented by $3 \cdot x + 3\cdot y + 2\cdot z = 6$ .

Step-by-step explanation:

The general equation in rectangular form for a 3-dimension plane is represented by:

$a\cdot x + b\cdot y + c\cdot z = d$

Where:

$x$ , $y$ , $z$ - Orthogonal inputs.

$a$ , $b$ , $c$ , $d$ - Plane constants.

The plane presented in the figure contains the following three points: (2, 0, 0), (0, 2, 0), (0, 0, 3)

For the determination of the resultant equation, three equations of line in three distinct planes orthogonal to each other. That is, expressions for the xy, yz and xz-planes with the resource of the general equation of the line:

xy-plane (2, 0, 0) and (0, 2, 0)

$y = m\cdot x + b$

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Where:

$m$ - Slope, dimensionless.

$x_{1}$ , $x_{2}$ - Initial and final values for the independent variable, dimensionless.

$y_{1}$ , $y_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - x-Intercept, dimensionless.

If $x_{1} = 2$ , $y_{1} = 0$ , $x_{2} = 0$ and $y_{2} = 2$ , then:

Slope

$m = \frac{2-0}{0-2}$

$m = -1$

x-Intercept

$b = y_{1} - m\cdot x_{1}$

$b = 0 -(-1)\cdot (2)$

$b = 2$

The equation of the line in the xy-plane is $y = -x+2$ or $x + y = 2$ , which is equivalent to $3\cdot x + 3\cdot y = 6$ .

yz-plane (0, 2, 0) and (0, 0, 3)

$z = m\cdot y + b$

$m = \frac{z_{2}-z_{1}}{y_{2}-y_{1}}$

Where:

$m$ - Slope, dimensionless.

$y_{1}$ , $y_{2}$ - Initial and final values for the independent variable, dimensionless.

$z_{1}$ , $z_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - y-Intercept, dimensionless.

If $y_{1} = 2$ , $z_{1} = 0$ , $y_{2} = 0$ and $z_{2} = 3$ , then:

Slope

$m = \frac{3-0}{0-2}$

$m = -\frac{3}{2}$

y-Intercept

$b = z_{1} - m\cdot y_{1}$

$b = 0 -\left(-\frac{3}{2} \right)\cdot (2)$

$b = 3$

The equation of the line in the yz-plane is $z = -\frac{3}{2}\cdot y+3$ or $3\cdot y + 2\cdot z = 6$ .

xz-plane (2, 0, 0) and (0, 0, 3)

$z = m\cdot x + b$

$m = \frac{z_{2}-z_{1}}{x_{2}-x_{1}}$

Where:

$m$ - Slope, dimensionless.

$x_{1}$ , $x_{2}$ - Initial and final values for the independent variable, dimensionless.

$z_{1}$ , $z_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - z-Intercept, dimensionless.

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Slope

$m = \frac{3-0}{0-2}$

$m = -\frac{3}{2}$

x-Intercept

$b = z_{1} - m\cdot x_{1}$

$b = 0 -\left(-\frac{3}{2} \right)\cdot (2)$

$b = 3$

The equation of the line in the xz-plane is $z = -\frac{3}{2}\cdot x+3$ or $3\cdot x + 2\cdot z = 6$

After comparing each equation of the line to the definition of the equation of the plane, the following coefficients are obtained:

$a = 3$ , $b = 3$ , $c = 2$ , $d = 6$

Hence, none of the options presented are valid. The plane is represented by $3 \cdot x + 3\cdot y + 2\cdot z = 6$ .

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Answer:

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Step-by-step explanation:

Given:

The sides of the triangle = 15 10 and 7

To Find:

Whether the triangle is a right triangle = ?

Solution:

According to Pythagorean theorem , In an right triangle the the sum of the squares of the two smaller sides must equal to the square of the larger side

$a^2 +b^2 =c^2\\$

where

a and b are the smaller sides

c is the larger side

Now from the given data, lets assume

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Substituting in the above equation we get,

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