All categories

2 years ago

Which is the pattern for factoring the sum of cubes

Mathematics

2 answers:

vovikov84 [41]2 years ago

4 0

Answer: Its B

Step-by-step explanation: trust me i had this test

Stels [109]2 years ago

3 0

Answer:

4th option

Step-by-step explanation:

The factors of a³ + b³ are

a³ + b³ = (a + b)(a² - ab + b² )

You might be interested in

Students are running in a relay race. Each team will run a total of 3 miles. Each member of a team will run 1/8 of a mile. How m

Crazy boy [7]

3 / (1/8) = 3 × (8/1) = 24
There needs to be 24 students on a team.

7 0

2 years ago

What does 5 2/3 - 8 5/6 equal?

goblinko [34]

5 2/3 = 17/3 = 34/6

8 5/6 = 53/6

34/6 - 53/6 = -19/6 = -3 1/6

7 0

3 years ago

Question 4) -6x + 9 + 6x -13<br> Your answer<br> Question 5) 8x - 9 -11 -14x<br> Your answer

Digiron [165]

Answer:

4) -4

5) -6x-20

Step-by-step explanation:

Add x with x's and numbers with numbers

4) -6x + 9 + 6x -13

-6x+6x+9-13

x's cancel

9-13 = -4

5) 8x - 9 -11 -14x

8x-14x-9-11

-6x-20

hope this helps

3 0

2 years ago

HELPPP PLSSSS!! SHOW YOUR WORK!!<br><br>THANKS

Kipish [7]

I belive it would be B. because of the fact that they are parallel which means their slope should stay the same...but their y-intersept would change

7 0

2 years ago

The quadratic function g(x) = a.ca + bx+c has the

Mumz [18]

<em>The value of b is 14 and the value of c is 65</em>

<h2>Explanation:</h2>

The quadratic function is a function of the form:

$f(x)=ax^2+bx+c$

Here we know that the leading coefficient $a=1$ so we reduce our equation to:

$g(x)=x^2+bx+c$

The roots are those values at which $g(x)=0$

So:

$x^2+bx+c=0 \\ \\ First \ root: \\ \\ (-7+4i)^2+b(-7+4i)+c=0 \\ \\ (-7)^2-2(7)(4i)+(4i)^2-7b+4bi+c=0 \\ \\ 49-56i+16i^2-7b+4bi+c=0 \\ \\ \\ Simplifying: \\ \\ 49-56i+16(-1)-7b+4bi+c=0 \\ \\ 49-56i-16-7b+4bi+c=0 \\ \\ 33-56i-7b+4bi+c=0 \\ \\ \\$

$Second \ root: \\ \\ (-7-4i)^2+b(-7-4i)+c=0 \\ \\ (-1)^2(7+4i)^2+b(-7-4i)+c=0 \\ \\ (7)^2+2(7)(4i)+(4i)^2-7b-4bi+c=0 \\ \\ 49+56i+16i^2-7b-4bi+c=0 \\ \\ \\ Simplifying: \\ \\ 49+56i+16(-1)-7b-4bi+c=0 \\ \\ 49+56i-16-7b-4bi+c=0 \\ \\ 33+56i-7b-4bi+c=0$

So we have:

$(1) \ 33-56i-7b+4bi+c=0 \\ \\ (2) \ 33+56i-7b-4bi+c=0 \\ \\ \\ Subtract \ 2 \ from: \\ \\ 33-56i-7b+4bi+c-(33+56i-7b-4bi+c)=0 \\ \\ 33-56i-7b+4bi+c-33-56i+7b+4bi-c=0 \\ \\ \\ Combine \ like \ terms: \\ \\ 33-33-56i-56i-7b+7b+4bi+4bi+c-c=0 \\ \\ -112i+8bi=0 \\ \\ Isolating \ b: \\ \\ b=\frac{112i}{8i} \\ \\ \boxed{b=14}$

Finding c from (1):

$33-56i-7b+4bi+c=0 \\ \\ \\ Substituting \ b: \\ \\ 33-56i-7(14)+4(14)i+c=0 \\ \\ 33-56i-98+56i+c=0 \\ \\ -65+c=0 \\ \\ \boxed{c=65}$

<h2>Learn more:</h2>

Complex conjugate: brainly.com/question/2137496

#LearnWithBrainly

5 0

3 years ago

Which is the pattern for factoring the sum of cubes​

Which is the pattern for factoring the sum of cubes