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3 years ago

I need to solve for n.

Mathematics

2 answers:

vaieri [72.5K]3 years ago

8 0

It’s going to be 286.

Arturiano [62]3 years ago

6 0

No ha ha dicho que no ha sido el caso de la capital de los 282

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If alpha and beta are the zeroes of the polynomial 6x2+x-2 find the value og alpha/beta + beta/alpha

castortr0y [4]

$6x^2+x-2=6x^2+4x-3x-2=2x(3x+2)-1(3x+2)\\\\=(3x+2)(2x-1)\\\\3x+2=0\to x=-\frac{2}{3}\\\\2x-1=0\to x=\frac{1}{2}\\\\\alpha=-\frac{2}{3};\ \beta=\frac{1}{2}\\\\\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{-\frac{2}{3}}{\frac{1}{2}}+\frac{\frac{1}{2}}{-\frac{2}{3}}=-\frac{2}{3}\cdot\frac{2}{1}-\frac{1}{2}\cdot\frac{3}{2}=-\frac{4}{3}-\frac{3}{4}\\\\=-\frac{16}{12}-\frac{9}{12}=-\frac{25}{12}=-2\frac{1}{12}$

$use\ Vieta's\ formula:\\\\\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{\alpha^2+2\alpha\beta+\beta^2-2\alpha\beta}{\alpha\beta}=\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}=\frac{(\alpha+\beta)^2}{\alpha\beta}-2\\\\\alpha+\beta=\frac{-b}{a};\ \alpha\beta=\frac{c}{a}\\\\\frac{(\alpha+\beta)^2}{\alpha\beta}-2=\frac{\left(\frac{-b}{a}\right)^2}{\frac{c}{a}}-2=\frac{b^2}{a^2}\cdot\frac{a}{c}-2=\frac{b^2}{ac}-2$

$6x^2+x-2\\\\a=6;\ b=1;\ c=-2\\\\\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{1^2}{6\cdot(-2)}-2=\frac{1}{-12}-2=-2\frac{1}{12}$

7 0

3 years ago

What is the answer please.... need help

mina [271]

Answer:

$f\left(x\right)=x^3-6x^2+3x+10$ is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Step-by-step explanation:

Given the function

$f\left(x\right)=x^3-6x^2+3x+10$

As the highest power of the x-variable is 3 with the leading coefficients of 1.

So, it is clear that the polynomial function of the least degree has the real coefficients and the leading coefficients of 1.

solving to get the zeros

$f\left(x\right)=x^3-6x^2+3x+10$

$0=x^3-6x^2+3x+10$ ∵ $f(x)=0$

$Factor\:x^3-6x^2+3x+10\::\:\left(x+1\right)\left(x-2\right)\left(x-5\right)=0$

$\left(x+1\right)\left(x-2\right)\left(x-5\right)=0$

Using the zero factor principle

if $ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

$x+1=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0$

$x=-1,\:x=2,\:x=5$

Therefore, the zeros of the function are:

$x=-1,\:x=2,\:x=5$

$f\left(x\right)=x^3-6x^2+3x+10$ is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Therefore, the last option is true.

8 0

3 years ago

Find the equation of the line that passes through points a and b

Blizzard [7]

Answer:

y - 7 = 2(x - 1)

Step-by-step explanation:

Going from (-3, -1) to (1, 7), x increases by 4 and y by 8. These numbers are the 'run' and 'rise' of the line, respectively. Thus, the slope of the red line is m = rise/run = m = 8/4 = 2.

Using the point-slope formula and the point (1, 7), we get:

y - 7 = 2(x - 1)

6 0

3 years ago

Will bought 3 college textbooks. One cost $32, one cost $45, and one cost $39. What is the average price of his books? Choose th

ch4aika [34]

I think the correct answer among the choices presented above is option B. Average is the sum of all the data divided the number of data added. From the statement, we have 3 prices to be added and divided by 3. Therefore, the average is calculated as

<span>($32 + $45 + $39) ÷ 3</span>

7 0

3 years ago

Those are the answer choices <br> I WILL GIVE BRAINLIEST!!

jek_recluse [69]

It’s the sixth one good luck stay safe and thank you if you give me brainliest

6 0

3 years ago

I need to solve for n.​

I need to solve for n.