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r-ruslan [8.4K]
4 years ago
9

When simplifying 4x^3-10x^2+6x/2x^3+x^2-3x, what are the term(s) that can be cancelled

Mathematics
1 answer:
brilliants [131]4 years ago
3 0

Answer:

x can be cancelled

Step-by-step explanation:

we are given

\frac{4x^3-10x^2+6x}{2x^3+x^2-3x}

Firstly, we will factor numerator and denominator

and then we can factor it

4x^3-10x^2+6x=2x(2x^2-5x+3)

4x^3-10x^2+6x=2x(2x+1)(x-3)

now, we can factor denominator

2x^3+x^2-3x=x(2x^2+x-3)

2x^3+x^2-3x=x(x-1)(2x+3)

now, we can replace it

\frac{4x^3-10x^2+6x}{2x^3+x^2-3x}=\frac{2x(2x+1)(x-3)}{x(x-1)(2x+3)}

we can see that

both terms are having x common

so, x can be cancelled

So,

x can be cancelled

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Answer:

Option B

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Step-by-step explanation:

we have

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we know that

If two system of equations are equivalent, then their solution is the same

Part 1)

step 1

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step 2

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5x-4y=-16 ----> equation b'

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The system of equations a and b and the system of equations a' and b' are equivalent

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The system of equations a' and b' have the same solution as the given system

Part 2)

step 1

Multiply by 5 equation a both sides

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The system of equations a and b and the system of equations a' and b' are equivalent

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Answer:

37 is the smallest positive integer n such that n(n+1)(n+2) is divisible by 247.

Step-by-step explanation:

First we will find the prime factors of 247:

247 = 13 x 19 (which are both prime).

So now we need to find a number (the smallest one) that is of the form (n)(n+1)(n+2) (the product of three consecutive numbers) and that is divisible by both 13 and 19 (and therefore divisible by 247)

Let's take a look at the multiples of 13: 13, 26, 39, 52...

Let's take a look at the multiples of 19: 19, 38, 57...

We can see that the first time we have two multiples close together are the 38 (for 19) and the 39 (for 17).

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