<img src="https://tex.z-dn.net/?f=%5Csqrt%5B6%5D%7B27x%5E%7B3%7Dy%5E%7B6%7D%20%20%7D" id="TexFormula1" title="\sqrt[6]{27x^{3}y^

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tia_tia [17]

3 years ago

{6} }" alt="\sqrt[6]{27x^{3}y^{6} }" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

bekas [8.4K]3 years ago

3 0

Answer:

$\large\boxed{\sqrt[6]{27x^3y^6}=y\sqrt{3x}}$

Step-by-step explanation:

$\sqrt[6]{27x^3y^6}\qquad\text{use}\ \sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b}\\\\=\sqrt[6]{3^3}\cdot\sqrt[6]{x^3}\cdot\sqrt[6]{y^6}\qquad\text{use}\ \sqrt[n]{a^m}=a^\frac{m}{n}\\\\=3^\frac{3}{6}\cdot x^\frac{3}{6}\cdot y^\frac{6}{6}\qquad\text{simplify}\\\\=3^\frac{1}{2}\cdot x^\frac{1}{2}\cdot y^1\qquad\text{use}\ (ab)^n=a^n\cdot b^n\\\\=(3x)^\frac{1}{2}\cdot y\\\\=y\sqrt{3x}$

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If the sum of the zereos of the quadratic polynomial is 3x^2-(3k-2)x-(k-6) is equal to the product of the zereos, then find k?

lys-0071 [83]

Answer:

Step-by-step explanation:

So I'm going to use vieta's formula.

Let u and v the zeros of the given quadratic in ax^2+bx+c form.

By vieta's formula:

1) u+v=-b/a

2) uv=c/a

We are also given not by the formula but by this problem:

3) u+v=uv

If we plug 1) and 2) into 3) we get:

-b/a=c/a

Multiply both sides by a:

-b=c

Here we have:

a=3

b=-(3k-2)

c=-(k-6)

So we are solving

-b=c for k:

3k-2=-(k-6)

Distribute:

3k-2=-k+6

Add k on both sides:

4k-2=6

Add 2 on both side:

4k=8

Divide both sides by 4:

k=2

Let's check:

$3x^2-(3k-2)x-(k-6) \text{ with }k=2$ :

$3x^2-(3\cdot 2-2)x-(2-6)$

$3x^2-4x+4$

I'm going to solve $3x^2-4x+4=0$ for x using the quadratic formula:

$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\frac{4\pm \sqrt{(-4)^2-4(3)(4)}}{2(3)}$

$\frac{4\pm \sqrt{16-16(3)}}{6}$

$\frac{4\pm \sqrt{16}\sqrt{1-(3)}}{6}$

$\frac{4\pm 4\sqrt{-2}}{6}$

$\frac{2\pm 2\sqrt{-2}}{3}$

$\frac{2\pm 2i\sqrt{2}}{3}$

Let's see if uv=u+v holds.

$uv=\frac{2+2i\sqrt{2}}{3} \cdot \frac{2-2i\sqrt{2}}{3}$

Keep in mind you are multiplying conjugates:

$uv=\frac{1}{9}(4-4i^2(2))$

$uv=\frac{1}{9}(4+4(2))$

$uv=\frac{12}{9}=\frac{4}{3}$

Let's see what u+v is now:

$u+v=\frac{2+2i\sqrt{2}}{3}+\frac{2-2i\sqrt{2}}{3}$

$u+v=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}$

We have confirmed uv=u+v for k=2.

4 0

3 years ago

A bicyclist rides 16 miles in 144 minutes. If she continues at this speed, how long will it take her to travel 36 miles?

mario62 [17]

16 -> 144
36 -> ?

(36x144) / 16 =324 min

6 0

3 years ago

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frez [133]

Answer: 125 or some thing

Step-by-step explanation:

5 0

3 years ago

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F(x)=x^2-1 <br> G(x)=x-1 <br> find (f+g)(10).

WARRIOR [948]

<h2>Adding Functions</h2><h3>Answer:</h3>

$(f +g)(10) = 108$

<h3>Step-by-step explanation:</h3>

Before we can solve for $(f +g)(10)$ , we need to know how $(f +g)(x)$ is defined.

$(f +g)(x) = f(x) +g(x) \\ (f +g)(x) = (x^2 -1) +(x -1) \\ (f +g)(x) = x^2 -1 +x -1 \\ (f +g)(x) = x^2 +x -2$

We can now solve for $(f +g)(10)$ :

$(f +g)(10) = (10)^2 +(10) -2 \\ (f +g)(10) = 100 +10 -2 \\ (f +g)(10) = 108$

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3 years ago

Which ordered pair is a solution to the system of linear equations 1/2x-3/4y=11/60 and 2/5x+1/6y=3/10

natka813 [3]

ANSWER

$( \frac{2}{3} , \frac{1}{5} )$

EXPLANATION

The first equation is

$\frac{1}{2} x - \frac{3}{4} y = \frac{11}{60} ...(1)$

The second equation is

$\frac{2}{5} x + \frac{1}{6} y = \frac{3}{10} ...(2)$

We want to eliminate y, so we multiply the first equation by

$\frac{4}{5}$

$\frac{4}{5} \times \frac{1}{2} x - \frac{4}{5} \times \frac{3}{4} y = \frac{11}{60} \times \frac{4}{5}$

$\frac{2}{5} x - \frac{3}{5} y = \frac{11}{75} ...(3)$

We now subtract equation (3) from (2)

$(\frac{2}{3} x - \frac{2}{3} x )+ ( \frac{1}{6} y - - \frac{3}{5}y ) =( \frac{3}{10} - \frac{11}{75} )$

$\frac{1}{6} y + \frac{3}{5}y =\frac{3}{10} - \frac{11}{75}$

$\frac{23}{30}y = \frac{23}{150}$

Multiply both sides by

$\frac{30}{23}$

$\implies \: \frac{30}{23} \times \frac{23}{30}y= \frac{23}{150} \times \frac{30}{23}$

$\implies \: y = \frac{1}{5}$

Substitute into the first equation to solve for x .

$\frac{1}{2} x - \frac{3}{4} \times \frac{1}{5} = \frac{11}{60}$

Multiply to obtain

$\frac{1}{2} x - \frac{3}{20} = \frac{11}{60}$

$\frac{1}{2} x = \frac{11}{60} + \frac{3}{20}$

$\frac{1}{2} x = \frac{1}{3}$

Multiply both sides by 2.

$2 \times \frac{1}{2} x =2 \times \frac{1}{3}$

$x = \frac{2}{3}$

The solution is

$( \frac{2}{3} , \frac{1}{5} )$

5 0

3 years ago

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