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dolphi86 [110]
3 years ago
13

Chad bought a box of raisins. he gave 1/6 of it to one brother, 1/4 to another brother, and he kept the rest for himself. what f

raction did he keep?
Mathematics
1 answer:
fenix001 [56]3 years ago
4 0
Greetings!

This can be solved by creating a simple equation.
1/4+1/6+x=1

Solve for x.
1/4+1/6+x=1
Find the LCM.
3/12+2/12+x=1
5/12+x=1
Add -5/12 to both sides.
(5/12+x)+(-5/12)=(1)+(-5/12)
Simplify.
x=1/1-5/12
Find the LCM.
x=12/12-5/12
x=7/12

He kept 7/12 of the raisins for himself.

Hope this helps.
-Benjamin
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For what values of x is f(x) = |x + 1| differentiable? I'm struggling my butt off for this course
pav-90 [236]

By definition of absolute value, you have

f(x) = |x+1| = \begin{cases}x+1&\text{if }x+1\ge0 \\ -(x+1)&\text{if }x+1

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f(x) = \begin{cases}x+1&\text{if }x\ge-1\\-x-1&\text{if }x

On their own, each piece is differentiable over their respective domains, except at the point where they split off.

For <em>x</em> > -1, we have

(<em>x</em> + 1)<em>'</em> = 1

while for <em>x</em> < -1,

(-<em>x</em> - 1)<em>'</em> = -1

More concisely,

f'(x) = \begin{cases}1&\text{if }x>-1\\-1&\text{if }x

Note the strict inequalities in the definition of <em>f '(x)</em>.

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\displaystyle \lim_{x\to-1^-}f'(x) = \lim_{x\to-1}(-1) = -1

\displaystyle \lim_{x\to-1^+}f'(x) = \lim_{x\to-1}1 = 1

All this to say that <em>f(x)</em> is differentiable everywhere on its domain, <em>except</em> at the point <em>x</em> = -1.

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