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3 years ago

Explain how to use a polynomial identity to factor x^6-27y^3

Mathematics

2 answers:

Umnica [9.8K]3 years ago

6 0

Dont really no sorry

Taya2010 [7]3 years ago

5 0

This is a sum of cubes:

x⁶ - 27y³ = (x²)³ + (-3y)³

Recall that

a³ + b³ = (a + b) (a² - ab + b²)

So we have

x⁶ - 27y³ = (x² - 3y) ((x²)² - x² (-3y) + (-3y)²)

x⁶ - 27y³ = (x² - 3y) (x⁴ + 3x²y + 9y²)

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What is vertex of x^2+5x-3

Art [367]

$\text{The vertex of}\ ax^2+bx+c:\\\\(h,\ k),\ \text{where}\ h=\dfrac{-b}{2a};\ k=f(h)=\dfrac{-(-b^2-4ac)}{4a}$

$\text{We have:}\\f(x)=x^2+5x-3\\\\a=1,\ b=5,\ c=-3\\\\h=\dfrac{-5}{2\cdot1}=\dfrac{-5}{2}=-2.5\\\\k=f(-2.5)=(-2.5)^2+5(-2.5)-3=6.25-12.5-3=-9.25$

$\text{Answer: the vertex is in}\ (-2.5;\ -9.25)$

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3 years ago

Alex earns 12% commission on any of her monthly sales over $5,000. Her sales for May were $6,750. How much commission did Alex e

valentina_108 [34]

6750/100*12 = 810 is the answer

4 0

3 years ago

Name a number whose square root and cube root are whole numbers

katrin [286]

The whole number is 10 the square root is 9

5 0

3 years ago

Alexei wants to hang a mirror in his boat and put a frame around it. The mirror and frame must have an area of 19.25 square feet

zhuklara [117]

Let

x---------> the thickness of the frame in feet

At-------> Area of the mirror and the frame in square feet

we know that

$At=19.25ft^{2}$ ------> equation $1$

$At=(3+2x)*(5+2x)\\ At=3*5+3*2x+5*2x+2x*2x\\ At=15+6x+10x+4x^{2} \\ At=4x^{2} +16x+15$ --------> equation $2$

equate equation $1$ and equation $2$

$19.25=4x^{2} +16x+15\\ 4x^{2} +16x+15-19.25=0\\ 4x^{2} +16x-4.25=0$

therefore

the answer is the option

$C. 4x2 + 16x - 4.25 = 0$

5 0

3 years ago

Read 2 more answers

Apply Crammer's Rule to find the solution to the following quations .2x + 3y = 1, 3x + y = 5

Bond [772]

Answer:

The solution to the equation system given is:

x = 2
y = -1

Step-by-step explanation:

First, we must know the equations given:

2x + 3y = 1
3x + y = 5

Following Crammer's Rule, we have the matrix form:

$\left[\begin{array}{ccc}2&3\\3&1\end{array}\right] =\left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}1\\5\end{array}\right]$

Now we solve using the determinants:

$x=\frac{\left[\begin{array}{ccc}1&3\\5&1\end{array}\right]}{\left[\begin{array}{ccc}2&3\\3&1\end{array}\right] } =\frac{(1*1)-(5*3)}{(2*1)-(3*3)} = \frac{1-15}{2-9} =\frac{-14}{-7} = 2$

$y=\frac{\left[\begin{array}{ccc}2&1\\3&5\end{array}\right]}{\left[\begin{array}{ccc}2&3\\3&1\end{array}\right] } =\frac{(2*5)-(3*1)}{(2*1)-(3*3)}=\frac{10-3}{2-9} =\frac{7}{-7}=-1$

Now, we can find the answer which is x= 2 and y= -1, we can replace these values in the equation to confirm the results are right, with the first equation:

2x + 3y = 1
2(2) + 3(-1)= 1
4 - 3 = 1
1 = 1

And, with the second equation:

3x + y = 5
3(2) + (-1) = 5
6 - 1 = 5
5 = 5

4 0

3 years ago