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Amiraneli [1.4K]

3 years ago

10

Can someone please help me, this is for summer school and I need fast replies, I have 29 other question like these, please help.

1 answer:

Yuri [45]3 years ago

7 0

$\implies {\blue {\boxed {\boxed {\purple {\sf { \: A. \:- 8 {v}^{4} + 3 {v}^{3} + v + 5}}}}}}$

$\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\red{:}}}}}$

$(4 + 6v + 3 {v}^{3} ) - (5v + 8 {v}^{4} - 1)$

➼ $\: 4 + 6v + 3 {v}^{3} - 5v - 8 {v}^{4} + 1$

Combining like terms, we have

➼ $\: - 8 {v}^{4} + 3 {v}^{3} + 6v - 5v + 4 + 1$

➼ $\: - 8 {v}^{4} + 3 {v}^{3} + v + 5$

<h3 />

$\large\mathfrak{{\pmb{\underline{\orange{Mystique35 }}{\orange{❦}}}}}$

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How to factor a tri, quad, or polynomial.

Akimi4 [234]

Explanation:

Factoring to linear factors generally involves finding the roots of the polynomial.

The two rules that are taught in Algebra courses for finding real roots of polynomials are ...

Descartes' rule of signs: the number of positive real roots is equal to the number of coefficient sign changes when the polynomial is written in standard form.
Rational root theorem: possible rational roots will have a numerator magnitude that is a divisor of the constant, and a denominator magnitude that is a divisor of the leading coefficient when the coefficients of the polynomial are rational. (Trial and error will narrow the selection.)

In general, it is a difficult problem to find irrational real factors, and even more difficult to find complex factors. The methods for finding complex factors are not generally taught in beginning Algebra courses, but may be taught in some numerical analysis courses.

Formulas exist for finding the roots of quadratic, cubic, and quartic polynomials. Above 2nd degree, they tend to be difficult to use, and may produce results that are less than easy to use. (The real roots of a cubic may be expressed in terms of cube roots of a complex number, for example.)

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Personally, I find a graphing calculator to be exceptionally useful for finding real roots. A suitable calculator can find irrational roots to calculator precision, and can use that capability to find a pair of complex roots if there is only one such pair.

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<em>Additional comment</em>

Some algebra courses teach iterative methods for finding real zeros. These can include secant methods, bisection, and Newton's method iteration. There are anomalous cases that make use of these methods somewhat difficult, but they generally can work well if an approximate root value can be found.

6 0

3 years ago

Identify the type of sampling method used.A trucking company places their office phone number on the back of all of their vehicl

ziro4ka [17]

Explanation

In this sample, people elect to participate or self-select to do the survey. Often, these folks have a strong interest in the main topic of the survey.

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7 0

2 years ago

Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.

mote1985 [20]

Answer:

$\frac{d}{dx}\left(\ln \left(\frac{x}{x^2+1}\right)\right)=\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\frac{-x^2+1}{x\left(x^2+1\right)}$

Step-by-step explanation:

To find the derivative of the function $y(x)=\ln \left(\frac{x}{x^2+1}\right)$ you must:

Step 1. Rewrite the logarithm:

$\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\left(\ln{\left(x \right)} - \ln{\left(x^{2} + 1 \right)}\right)^{\prime }$

Step 2. The derivative of a sum is the sum of derivatives:

$\left(\ln{\left(x \right)} - \ln{\left(x^{2} + 1 \right)}\right)^{\prime }}={\left(\left(\ln{\left(x \right)}\right)^{\prime } - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }\right)$

Step 3. The derivative of natural logarithm is $\left(\ln{\left(x \right)}\right)^{\prime }=\frac{1}{x}$

${\left(\ln{\left(x \right)}\right)^{\prime }} - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }={\frac{1}{x}} - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }$

Step 4. The function $\ln{\left(x^{2} + 1 \right)}$ is the composition $f\left(g\left(x\right)\right)$ of two functions $f\left(u\right)=\ln{\left(u \right)}$ and $u=g\left(x\right)=x^{2} + 1$

Step 5. Apply the chain rule $\left(f\left(g\left(x\right)\right)\right)^{\prime }=\frac{d}{du}\left(f\left(u\right)\right) \cdot \left(g\left(x\right)\right)^{\prime }$

$-{\left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }} + \frac{1}{x}=- {\frac{d}{du}\left(\ln{\left(u \right)}\right) \frac{d}{dx}\left(x^{2} + 1\right)} + \frac{1}{x}\\\\- {\frac{d}{du}\left(\ln{\left(u \right)}\right)} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}=- {\frac{1}{u}} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}$

Return to the old variable:

$- \frac{1}{{u}} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}=- \frac{\frac{d}{dx}\left(x^{2} + 1\right)}{{\left(x^{2} + 1\right)}} + \frac{1}{x}$

The derivative of a sum is the sum of derivatives:

$- \frac{{\frac{d}{dx}\left(x^{2} + 1\right)}}{x^{2} + 1} + \frac{1}{x}=- \frac{{\left(\frac{d}{dx}\left(1\right) + \frac{d}{dx}\left(x^{2}\right)\right)}}{x^{2} + 1} + \frac{1}{x}=\frac{1}{x^{3} + x} \left(x^{2} - x \left(\frac{d}{dx}\left(1\right) + \frac{d}{dx}\left(x^{2}\right)\right) + 1\right)$

Step 6. Apply the power rule $\frac{d}{dx}\left(x^{n}\right)=n\cdot x^{-1+n}$

$\frac{1}{x^{3} + x} \left(x^{2} - x \left({\frac{d}{dx}\left(x^{2}\right)} + \frac{d}{dx}\left(1\right)\right) + 1\right)=\\\\\frac{1}{x^{3} + x} \left(x^{2} - x \left({\left(2 x^{-1 + 2}\right)} + \frac{d}{dx}\left(1\right)\right) + 1\right)=\\\\\frac{1}{x^{3} + x} \left(- x^{2} - x \frac{d}{dx}\left(1\right) + 1\right)\\$

$\frac{1}{x^{3} + x} \left(- x^{2} - x {\frac{d}{dx}\left(1\right)} + 1\right)=\\\\\frac{1}{x^{3} + x} \left(- x^{2} - x {\left(0\right)} + 1\right)=\\\\\frac{1 - x^{2}}{x \left(x^{2} + 1\right)}$

Thus, $\frac{d}{dx}\left(\ln \left(\frac{x}{x^2+1}\right)\right)=\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\frac{-x^2+1}{x\left(x^2+1\right)}$

3 0

3 years ago

2 1/2 divide by 1/4 and how to write a real problem for it

den301095 [7]

2 1/2 = (2*2+1)/2 = 5/2
5/2 : 1/4 = 5/2 * 4/1 = 20/2 = 10

5 0

4 years ago

Julio bought 3 bags of cat food and 4 bags of dog food for $43.00 Sandy bought 3 bags of cat foot and 6 bags of dog food for $54

aleksklad [387]

Answer:

$12.50

Step-by-step explanation:

c is cat food d is dog food

3c + 4d = 43

3c + 6d = 54

subtract those to get

2d = 11

d = 5.50

3c + 2d = 3c + 22 = 43

3c = 21

c = 7

c+d = 7 + 5.50 = 12.50

7 0

3 years ago