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mash [69]
3 years ago
10

Please solve: m²j³j⁵/m⁸j⁻²m³

Mathematics
1 answer:
Alecsey [184]3 years ago
7 0

Step-by-step explanation:

first, simplify each part.

m²j^8

m¹¹j^-2

now put them together:

(m²j^8)÷(m¹¹j^-2)=m^-9 j^10

if anything is unclear, ask freely.

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Factored 8x2y3-5= <br> I need the answer
ANTONII [103]

Answer: Reformatting the input :

Changes made to your input should not affect the solution:

(1): "y3" was replaced by "y^3". 1 more similar replacement(s).

STEP

1

:

Equation at the end of step 1

(23x2 • y3) - 5

STEP

2

:

Trying to factor as a Difference of Cubes

2.1 Factoring: 8x2y3-5

Theory : A difference of two perfect cubes, a3 - b3 can be factored into

(a-b) • (a2 +ab +b2)

Proof : (a-b)•(a2+ab+b2) =

a3+a2b+ab2-ba2-b2a-b3 =

a3+(a2b-ba2)+(ab2-b2a)-b3 =

a3+0+0+b3 =

a3+b3

Check : 8 is the cube of 2

Check : 5 is not a cube !!

Ruling : Binomial can not be factored as the difference of two perfect cubes

Final result :

8x2y3 - 5

5 0
3 years ago
For the high school basketball game, it costs $6 for every 2 tickets. Complete the table below showing the cost and the number o
lara31 [8.8K]

Answer:

1. 12

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Step-by-step explanation:

1 ticket is $3

7 0
3 years ago
Read 2 more answers
The fraction a/b is in simplest terms and equal to the following:((2020^3)-1)/((2021^3)+1) . What is a + b?
blagie [28]

Consider the expression

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Factorize the numerator and denominator a difference/sum of cubes:

\dfrac{(x-1)(x^2+x+1)}{((x+1)+1)((x+1)^2-(x+1)+1)}

Expand the denominator:

\dfrac{(x-1)(x^2+x+1)}{(x+2)((x^2+2x+1)-(x+1)+1)}=\dfrac{(x-1)(x^2+x+1)}{(x+2)(x^2+x+1)}=\dfrac{x-1}{x+2}

since <em>x</em> = 2020, and clearly 2020² + 2020 + 1 ≠ 0, so we can cancel the factor of <em>x</em> ² + <em>x</em> + 1. This leaves us with

\dfrac{2020^3-1}{2021^3+1} = \dfrac{2020-1}{2020+2} = \dfrac{2019}{2022}=\dfrac{673}{674}=\dfrac ab

so that <em>a</em> + <em>b</em> = 673 + 674 = 1347.

7 0
3 years ago
Does the rule y= 2x^2 represent a expontional function
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Answer:

Yes

Step-by-step explanation:

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