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Mazyrski [523]
2 years ago
11

Simplify a(cube)-1000b(cube) 64a(cube)-125b(cube)

Mathematics
1 answer:
Elanso [62]2 years ago
8 0

The simplification of a³ - 1000b³ and 64a³ - 125b³ is (a - 10b) × (a² + 10ab + 100b²) and 4a - 5b) • (16a² + 20ab + 25b²) respectively.

<h3>Simplification</h3>

Question 1: a³ - 1000b³

a³ - b³

= (a-b) × (a² +ab +b²)

  • 1000 is the cube of 10
  • a³ is the cube of a¹
  • b³ is the cube of b¹

So,

(a - 10b) × (a² + 10ab + 100b²)

Question 2: 64a³ - 125b³

a³ - b³

= (a-b) × (a² +ab +b²)

  • 64 is the cube of 4
  • 125 is the cube of 5
  • a³ is the cube of a¹
  • b³ is the cube of b¹

So,

(4a - 5b) • (16a² + 20ab + 25b²)

Learn more about simplification:

brainly.com/question/723406

#SPJ1

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Answer:

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Step-by-step explanation:

4[24-3(7x+6)]-6(4-14x)

4(24-21x-18)-24+84x

96-84x-72-24+84x

96-72-24-84x+84x

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Write the explicit formula for the geometric sequence. Then find the fifth term in the sequence.
Blababa [14]
Sequence: -4, 8, -16

rate of increase: -16/8 = -2
 
If you make A1 the first term in the sequence you must use (n-1) in the explicit formula. If you go backwards in the sequence to find A0 = 2 then you would use n.

The don't have A0 = 2 listed as an option so we use A1 = -4 and (n-1) terms.

An = -4(-2)^(n-1)

Solving for the 5th term A5, use n = 5

A5 = -4(-2)^(5-1)
A5 = -4(16)
A5 = -64

So the answer is B.

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The area of a square is 64n^36 square units. what is the side length of one side of the square
sesenic [268]
Area = (side length)^2

Here,
 64n^36 = s^2.  Take the sqrt of both sides, obtaining 8n^18 = s.  This is the principal square root.  We use this positive quantity because the measure of side length must be positive.

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What is the value of a in the equation 3a + b = 54, when b = 9?
sergey [27]
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Match the system of equations on the left with the number of solutions on the right
erastova [34]

Answer:

top to bottom, the answers are b, c, a

Step-by-step explanation:

One way to find the solution to a system of equations is to substitute values in. For the first one,

y=2x+3

y=2x+5,

we can substitute 2x+3 =y into the second equation to get

y=2x+5

2x+3 = 2x+5

subtract 2x from both sides

3 = 5

As 3 is not equal to 5, this is never equal and therefore has no solution

For the second one,

y= 2x+7

y = (-2/3)x + 10

We can plug y=2x+7 into the second equation to get

2x + 7 = y = (-2/3)x + 10

2x + 7 = (-2/3)x + 10

add (2/3)x to both sides to make all x values on one side

2x + (2/3)x + 7  = 10

subtract 7 from both sides to make only x values on one side and only constants on the other

2x + (2/3)x = 3

(6/3)x + (2/3)x = 3

(8/3)x = 3

multiply both sides by 3 to remove a denominator

8x = 9

divide both sides by 8 to isolate x

x=9/8

There is only one value for when the equations are equal, so this has one solution

For the third one

y = x-5

2y = 2x - 10

Plug x-5 = y into the second equation

2 * y= 2*(x-5)

2 * (x-5) = 2x - 10

2x-10 = 2x-10

add 10 to both sides

2x=2x

As 2x is always equal to 2x, no matter what x is, there are infinitely many solutions for this system

6 0
3 years ago
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