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jok3333 [9.3K]
3 years ago
5

Multiply or divide. Show your work 5x+10 . 2x _____ ___ x+2 4x-10

Mathematics
1 answer:
sp2606 [1]3 years ago
8 0
\frac{2x(5x+10)}{(x+2)(4x-10)} = \frac{10x(x+2)}{2(x+2)(2x-5)} = \frac{5x}{2x+5}
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If x^2+y^2=1, what is the largest possible value of |x|+|y|?
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If <em>x</em>² + <em>y</em>² = 1, then <em>y</em> = ±√(1 - <em>x</em>²).

Let <em>f(x)</em> = |<em>x</em>| + |±√(1 - <em>x</em>²)| = |<em>x</em>| + √(1 - <em>x</em>²).

If <em>x</em> < 0, we have |<em>x</em>| = -<em>x</em> ; otherwise, if <em>x</em> ≥ 0, then |<em>x</em>| = <em>x</em>.

• Case 1: suppose <em>x</em> < 0. Then

<em>f(x)</em> = -<em>x</em> + √(1 - <em>x</em>²)

<em>f'(x)</em> = -1 - <em>x</em>/√(1 - <em>x</em>²) = 0   →   <em>x</em> = -1/√2   →   <em>y</em> = ±1/√2

• Case 2: suppose <em>x</em> ≥ 0. Then

<em>f(x)</em> = <em>x</em> + √(1 - <em>x</em>²)

<em>f'(x)</em> = 1 - <em>x</em>/√(1 - <em>x</em>²) = 0   →   <em>x</em> = 1/√2   →   <em>y</em> = ±1/√2

In either case, |<em>x</em>| = |<em>y</em>| = 1/√2, so the maximum value of their sum is 2/√2 = √2.

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