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vodka [1.7K]
3 years ago
14

Sara cut off 1/8 of a 48-inch piece of ribbon. Jane cut off 1/9 of a 36-inch piece of ribbon. They compared their cut pieces. Wh

ose piece is longer? How much longer? 
Mathematics
2 answers:
Vaselesa [24]3 years ago
7 0
Sara cut biggert part, and her ribbon was longer.

\frac{1}{8} * 48= \frac{48}{8} =6 \\\frac{1}{9} * 36= \frac{9}{36} =4 \\6>4 \\6-4=2
Sara's piece was longer by 2 inches.

Scrat [10]3 years ago
7 0

Answer:

Sara's by one whole

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Answer:

The maximum acceleration over that interval is A(6) = 28.

Step-by-step explanation:

The acceleration of this car is modelled as a function of the variable t.

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Apply the power rule to find the first and second derivatives of A(t):

\begin{aligned} A^{\prime}(t) &= 3\, t^{2} - 15\, t + 12 \\ &= 3\, (t - 1) \, (t + 4)\end{aligned}.

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Notice that both t = 1 and t = 4 are first derivatives of A^{\prime}(t) over the interval 0 \le t \le 6.

However, among these two zeros, only t = 1\! ensures that the second derivative A^{\prime\prime}(t) is smaller than zero (that is: A^{\prime\prime}(1) < 0.) If the second derivative A^{\prime\prime}(t)\! is non-negative, that zero of A^{\prime}(t) would either be an inflection point (ifA^{\prime\prime}(t) = 0) or a local minimum (if A^{\prime\prime}(t) > 0.)

Therefore \! t = 1 would be the only local maximum over the interval 0 \le t \le 6\!.

Calculate the value of A(t) at this local maximum:

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Compare these three possible maximum values of A(t) over the interval 0 \le t \le 6. Apparently, t = 6 would maximize the value of A(t)\!. That is: A(6) = 28 gives the maximum value of \! A(t) over the interval 0 \le t \le 6\!.

However, note that the maximum over this interval exists because t = 6\! is indeed part of the 0 \le t \le 6 interval. For example, the same A(t) would have no maximum over the interval 0 \le t < 6 (which does not include t = 6.)

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