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docker41 [41]
3 years ago
11

(2x + 1)(3x - 2) /24 x^2 - 4X - 8

Mathematics
1 answer:
iren2701 [21]3 years ago
8 0
I'm gonna go ahead and assume you're asking me to solve ((2x+1)(3x-2))/(24x^2-4x-8). 
((2x+1)(3x-2))/(24x^2-4x-8)
1. Factor out the common term 4. 
((2x+1)(3x-2))/(4(6x^2-x-2))
2. Split the second term in 6x^2-x-2 into two terms. 
((2x+1)(3x-2))/(4(6x+3x-4x-2))
3. Factor out common terms in the first two terms, then in the last two terms. 
((2x+1)(3x-2))/(4(3x(2x+1)-2(2x+1)))
4. Factor out the common term 2x+1
((2x+1)(3x-2))/(4(2x+1)(3x-2))
5. Cancel 3x-2
1/4
The answer to your question is 1/4. 
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Answer: Irrational

I'm going to write 'x' in place of "square root of 2" just to save time. Let's consider the possibility that 7-x is rational. I'm going to show there's a contradiction which ultimately concludes that the expression is irrational.

If 7-x was rational, then 7-x = p/q for some integers p,q with q being nonzero. Solve for x to get

7-x = p/q

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x = -(p/q) + 7

x = 7 - (p/q)

x = (7q/q) - (p/q)

x = (7q-p)/q

So if 7-x were rational, then this forces x to be rational because 7q-p is an integer over q (also an integer). The format is rational = integer/integer. But this is where the contradiction lies. Remember that x is standing in place for "square root of 2", which is an irrational number. There is no way to write sqrt(2) as a ratio of two integers. So x cannot be both rational and irrational at the same time.

Therefore, 7-x must be irrational making 7-sqrt(2) to be irrational as well.

note: sqrt is shorthand for "square root"

another note: adding any rational number to an irrational number will always result in an irrational number.

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If a angle is 4 times bigger than its supplement what will its measure be
Ne4ueva [31]
Supplementary angles mean that two angles equal to 180°.

Let us assume the 2 angles are angle A and angle B.

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Let Angle B be represented by x.

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4x + x = 180°
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Step-by-step explanation:

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