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Hatshy [7]
3 years ago
10

What is the value of x, when 10(x + 2) = 5(x + 8)?

Mathematics
1 answer:
tatuchka [14]3 years ago
7 0
The value of x is 4. The steps to solving the equation are to use distributive property first so that the equation then looks like this: 10x+20=5x+40
Next you would combine like terms and then the equation would look like this: 5x+20=40
Then you would use the subtraction property of equality to subtract 20 from 40 so your equation will now look like this: 5x=20
Then you use the division property of equality to divide 20 by 5 to get x=4
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3 years ago
If f(x)=x^3-x+2, then (f^-1)'(2)
yawa3891 [41]

Note that f(x) as given is <em>not</em> invertible. By definition of inverse function,

f\left(f^{-1}(x)\right) = x

\implies f^{-1}(x)^3 - f^{-1}(x) + 2 = x

which is a cubic polynomial in f^{-1}(x) with three distinct roots, so we could have three possible inverses, each valid over a subset of the domain of f(x).

Choose one of these inverses by restricting the domain of f(x) accordingly. Since a polynomial is monotonic between its extrema, we can determine where f(x) has its critical/turning points, then split the real line at these points.

f'(x) = 3x² - 1 = 0   ⇒   x = ±1/√3

So, we have three subsets over which f(x) can be considered invertible.

• (-∞, -1/√3)

• (-1/√3, 1/√3)

• (1/√3, ∞)

By the inverse function theorem,

\left(f^{-1}\right)'(b) = \dfrac1{f'(a)}

where f(a) = b.

Solve f(x) = 2 for x :

x³ - x + 2 = 2

x³ - x = 0

x (x² - 1) = 0

x (x - 1) (x + 1) = 0

x = 0   or   x = 1   or   x = -1

Then \left(f^{-1}\right)'(2) can be one of

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• 1/f'(0) = -1, if we restrict to (-1/√3, 1/√3); or

• 1/f'(1) = 1/2, if we restrict to (1/√3, ∞)

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2 years ago
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Step-by-step explanation:

Q 2/3 ÷ 3/7

1 change the divison sign

x = ÷

2 then the second fraction flip it

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3 then multiply the fraction

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