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3 years ago

Factor the expression: x^6-729. Show your work Please.

Mathematics

1 answer:

Brut [27]3 years ago

7 0

Answer:

(x + 3)(x - 3)(x^2 -3x + 9)(x^2 + 3x + 9)

Step-by-step explanation:

Use of formulas:

a^2 - b^2 = (a + b)(a -b)
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
a^3 - b^3 = (a - b)(a^2 + ab + b^2)

Given the expression:

x^6-729

Factoring 729

729 = 3*3*3*3*3*3 = 3^6

Factoring the expression

x^6 - 3^6 =
(x^3)^2 - (3^3)^2 =
(x^3 + 3^3)(x^3 - 3^3) =
(x + 3)(x^2 -3x + 9)(x - 3)(x^2 + 3x + 9) =
(x + 3)(x - 3)(x^2 -3x + 9)(x^2 + 3x + 9)

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what is the equation of the line that passes through the points (-2,1) and (1,10)? a) 3x-y=-7 b)3x-y=-5 c)3x-y=5 d)x 3y=-5

Ber [7]

The answer is a.
Substitute (-2,1) and (1,10) into each answer. Only a is correct because you will get -7 from both points.

8 0

3 years ago

#14 quick answer pls :)

Andrej [43]

Answer:

35.56

Step-by-step explanation:

4 0

2 years ago

First make a substitution and then use integration by parts to evaluate the integral. (Use C for the constant of integration.) x

e-lub [12.9K]

Answer:

$(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C$

Step-by-step explanation:

Ok, so we start by setting the integral up. The integral we need to solve is:

$\int x ln(5+x)dx$

so according to the instructions of the problem, we need to start by using some substitution. The substitution will be done as follows:

U=5+x

du=dx

x=U-5

so when substituting the integral will look like this:

$\int (U-5) ln(U)dU$

now we can go ahead and integrate by parts, remember the integration by parts formula looks like this:

$\int (pq')=pq-\int qp'$

so we must define p, q, p' and q':

p=ln U

$p'=\frac{1}{U}dU$

$q=\frac{U^{2}}{2}-5U$

q'=U-5

and now we plug these into the formula:

$\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int \frac{\frac{U^{2}}{2}-5U}{U}dU$

Which simplifies to:

$\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int (\frac{U}{2}-5)dU$

Which solves to:

$\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\frac{U^{2}}{4}+5U+C$

so we can substitute U back, so we get:

$\int xln(x+5)dU=(\frac{(x+5)^{2}}{2}-5(x+5))ln(x+5)-\frac{(x+5)^{2}}{4}+5(x+5)+C$

and now we can simplify:

$\int xln(x+5)dU=(\frac{x^{2}}{2}+5x+\frac{25}{2}-25-5x)ln(5+x)-\frac{x^{2}+10x+25}{4}+25+5x+C$

$\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}-\frac{5x}{2}-\frac{25}{4}+25+5x+C$

$\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C$

notice how all the constants were combined into one big constant C.

7 0

3 years ago

What is the image point of (1, -2) after a translation right 4 units and down 4 units?

zaharov [31]

It would be 4 over -4

3 0

2 years ago

PLEASE SHOW YOUR WORK.

schepotkina [342]

Answer:

a) Shawn's error was that he Multiplied 15 by 2x only. He didn't Multiply 15 by 7

b) The difference, in square feet, between the actual area of Shawn’s garden and the area found using his expression is given as

98 square feet

Step-by-step explanation:

The area in, square feet, of Shawn’s garden is found be calculating 15(2x + 7). Shawn incorrectly says the area can also be found using the expression 30x + 7.

The correct area =

15(2x + 7).

= 30x + 105 square feet

The error in Shawn’s expression is

= 15(2x + 7)

= 30x + 7 square feet

Shawn's error was that he Multiplied 15 by 2x only. He didn't Multiply 15 by 7

The difference, in square feet, between the actual area of Shawn’s garden and the area found using his expression is given as

30x + 105 square feet - 30x + 7 square feet

= 30x + 105 - (30x + 7)

= 30x - 30x + 105 - 7

= 98 square feet

7 0

2 years ago