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Alenkinab [10]
3 years ago
7

lou thinks the expression 27x^3-64 factors to (3x-4) (9x^2+12x+16). is he correct? How can you check his work? Show all work to

justify your thinking.
Mathematics
1 answer:
larisa [96]3 years ago
4 0

Answer:

Step-by-step explanation:

(a³ - b³) = (a - b) (a² + ab + b²)

27x³ - 64 = 3³x³ - 4³

               = (3x)³ - 4³   {a = 3x & b= 4)

Substitute in the above formula.

               = (3x -4)([3x]² + 3x*4 + 4²)

              = (3x - 4)(9x² + 12x + 16)

The expression 27x^3-64 factors to (3x-4) (9x^2+12x+16).

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Answer:

-n+m

Step-by-step explanation:

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Which numbers in set ={,,,,,,,,,}eh equals left brace 1 comma 2 comma 3 comma 4 comma 5 comma 6 comma 7 comma 8 comma 9 comma 10
viva [34]

Answer:

D={3,9}

The numbers are 3 and 9

Step-by-step explanation:

The set A

={,,,,,,,,,}

Let B be the sub set of A containing odd numbers

B={1,3,5,7,9}

Let C be the sub set of A containing multiply of 3

C= {3,6,9}

Now let D be the be the sub set of A containing both odd numbers and multiples of 3

D={3,9}

5 0
3 years ago
If 3 is subtracted from twice a number, the result is 8 less than the number. Write an equation to solve this problem.
lakkis [162]
3 subtracted from twice a number=2x-3
8 less than the number=x-8
2x-3=x-8
x=-5
3 0
3 years ago
First time on Brainly! Please Help!!!!! &gt;^&lt;
lys-0071 [83]
Welcome to Brainly!

Binomial Theorem will follow this pattern:
The powers start at 3 and decrease down to 0 for the first term,
and start at 0 and increase to 3 for the other term.

(2x+4)^3 =

\rm \_\_(2x)^34^0+\_\_(2x)^24^1+\_\_(2x)^14^2+\_\_(2x)^04^3

See how the power counts down on the (2x)
and counts up on the 4?
That's the pattern that our expansion must follow.

I left a little space in front of each term.
The coefficient in front of each term will come from the fourth row of Pascal's Triangle. The one that looks like this:

1 3 3 1

Those are the coefficients we want:

\rm 1(2x)^34^0+3(2x)^24^1+3(2x)^14^2+1(2x)^04^3

We can clean this up a little bit by getting rid of some of the junk. Anything to the 0th power is 1. So let's suppress all of our 1's because multiplying by 1 is not important.

\rm (2x)^3+3(2x)^24+3(2x)4^2+4^3

Now apply exponent rule, distributing the power to both the 2 and the x where applicable.

\rm 2^3x^3+3\cdot2^2x^24+3\cdot2x4^2+4^3

Remember, multiplication is COMMUTATIVE, meaning we can multiply things in any order. So let's bring the numerical portion to the front of each term and multiply it all out.

\rm 2^3x^3+3\cdot2^2\cdot4x^2+3\cdot2\cdot4^2x+4^3
\rm 8x^3+48x^2+96x+64
6 0
3 years ago
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