All categories

2 years ago

How would the expression x^3-64 be written using difference of cubes

Mathematics

1 answer:

Alexeev081 [22]2 years ago

7 0

The expression x³ - 64 can be rewritten using difference of cubes is; (x - 4)(x² + 4x + 16)

<h3>How to write an expression as difference of cubes?</h3>

Difference of cubes can be rewritten in the form of;

a³ – b³ = (a – b)(a² + ab + b²)

So in this question, we are given;

x³ - 64 = x³ - (4)³

where a = x and b = 4

This gives us;

(x - 4)(x² + 4x + 4²)

⇒ (x - 4)(x² + 4x + 16)

Read more about difference in cubes at; brainly.com/question/2747971

#SPJ1

You might be interested in

7-4[3-(4y-5)] plz help me

ad-work [718]

Let's simplify step-by-step. 7−4(3−(4y−5))

Distribute: =7+(−4)(3)+(−4)(−4y)+(−4)(5)=7+−12+16y+−20

Combine Like Terms: =7+−12+16y+−20=(16y)+(7+−12+−20)=16y+−25

Answer: =16y−25

7 0

3 years ago

Read 2 more answers

Help with homework plzzzzz!!!!!!!!!!!!!

Zanzabum

I'm not good with intigers, but if you added two negative intigers together wouldn't it still be negative? If you added -1 to -2 you'd get -3. It would always be less than the two original integers.

7 0

3 years ago

A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has fa

gayaneshka [121]

Answer:

the probability the die chosen was green is 0.9

Step-by-step explanation:

Given that:

A bag contains two six-sided dice: one red, one green.

The red die has faces numbered 1, 2, 3, 4, 5, and 6.

The green die has faces numbered 1, 2, 3, 4, 4, and 4.

From above, the probability of obtaining 4 in a single throw of a fair die is:

P (4 | red dice) = $\dfrac{1}{6}$

P (4 | green dice) = $\dfrac{3}{6}$ = $\dfrac{1}{2}$

A die is selected at random and rolled four times.

As the die is selected randomly; the probability of the first die must be equal to the probability of the second die = $\dfrac{1}{2}$

The probability of two 1's and two 4's in the first dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^4$

= $\dfrac{4!}{2!(4-2)!} ( \dfrac{1}{6})^4$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^4$

= $6 \times ( \dfrac{1}{6})^4$

= $(\dfrac{1}{6})^3$

= $\dfrac{1}{216}$

The probability of two 1's and two 4's in the second dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $6 \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $( \dfrac{1}{6}) \times ( \dfrac{3}{6})^2$

= $\dfrac{9}{216}$

∴

The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's in both die = $\dfrac{1}{216} \times \dfrac{1}{2} + \dfrac{9}{216} \times \dfrac{1}{2}$

The probability of two 1's and two 4's in both die = $\dfrac{1}{432} + \dfrac{1}{48}$

The probability of two 1's and two 4's in both die = $\dfrac{5}{216}$

By applying Bayes Theorem; the probability that the die was green can be calculated as:

P(second die (green) | two 1's and two 4's ) = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)

P(second die (green) | two 1's and two 4's ) = $\dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}$