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Rama09 [41]
3 years ago
7

What is StartFraction 11 Over 12 EndFraction divided by one-third? A fraction bar labeled 1. Under the 1 are 3 boxes labeled one

-third. Under the 3 boxes are 4 boxes containing one-fourth. Under the 4 boxes are 12 boxes containing StartFraction 1 Over 12 EndFraction.
Mathematics
2 answers:
Sunny_sXe [5.5K]3 years ago
6 0

Answer:

b. 2 3/4

Step-by-step explanation:

igor_vitrenko [27]3 years ago
3 0

Answer:

11/12 ÷ 1/3 = 2 and three-fourths

11/12 ÷ 1/3 = 2 3/4

Step-by-step explanation:

What is StartFraction 11 Over 12 EndFraction divided by one-third? A fraction bar labeled 1. Under the 1 are 3 boxes labeled one-third. Under the 3 boxes are 4 boxes containing one-fourth. Under the 4 boxes are 12 boxes containing StartFraction 1 Over 12 EndFraction. 2 and one-fourth 2 and three-fourths 3 and one-third 3 and two-thirds

Given:

11/12 ÷ 1/3

= 11/12 × 3/1

= 33 / 12

= 2 9/12

= 2 3/4

= 2 and three-fourths

11/12 ÷ 1/3 = 2 3/4

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Read 2 more answers
How to take derivative of absolute value.
zhenek [66]

The absolute value function is defined as

|x| = \begin{cases}x & \text{for }x \ge 0 \\ -x & \text{for }x < 0\end{cases}

If x is strictly positive (x > 0), then |x| = x, and d|x|/dx = dx/dx = 1.

If x is strictly negative (x < 0), then |x| = -x, and d|x|/dx = d(-x)/dx = -1.

But if x = 0, the derivative doesn't exist!

In order for the derivative of a function f(x) to exist at x = c, the limit

\displaystyle \lim_{x\to c}\frac{f(x) - f(c)}{x-c}

must exist. This limit does not exist for f(x) = |x| and c = 0 because the value of the limit depends on which way x approaches 0.

If x approaches 0 from below (so x < 0), we have

\displaystyle \lim_{x\to 0^-}\frac{|x|}x = \lim_{x\to0^-}-\frac xx = -1

whereas if x approaches 0 from above (so x > 0), we have

\displaystyle \lim_{x\to 0^+}\frac{|x|}x = \lim_{x\to0^+}\frac xx = 1

But 1 ≠ -1, so the limit and hence derivative doesn't exist at x = 0.

Putting everything together, you can define the derivative of |x| as

\dfrac{d|x|}{dx} =  \begin{cases}1 & \text{for } x > 0 \\ \text{unde fined} & \text{for }x = 0 \\ -1  & \text{for }x < 0 \end{cases}

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3 years ago
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