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stellarik [79]
3 years ago
13

Jack’s favorite sports drink comes in a 20-ounce bottle. the manufacturer of the sports drink requires a tolerance of 0.4 ounce.

this means as long as the bottle is within 0.4 ounce of 20 ounces, the bottle can be sent to stores to be sold.
write an absolute value inequality that describes the acceptable weights, w, of a 20-ounce bottle of sports drink.

answer the question with an absolute value inequality.
Mathematics
1 answer:
Scrat [10]3 years ago
4 0

Answer:

Answer:

Step-by-step explanation:

The manufacturer requires that the weight differs from 20 oz in at most 0.4 oz, therefore we can write this difference (which can be either above or below 20 oz, to be smaller than or equal to 0.4 oz.

The weight can be:

1) smaller than or equal to 20 oz + 0.4 oz:    then  

2) larger than or equal to 20 oz - 0.4 oz:    then  

and which combined, can be written as a double inequality;

This double condition can also be written using the absolute value symbol as:  

Step-by-step explanation:

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An increased the quantities of all the ingredients in a recipe by 60, percent. She used 80 grams left parenthesis, right parenth
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8 0
3 years ago
Find (a) the arc length and (b) the area of a sector.
Brut [27]

Answer:

a) 23.56 ft (2 dp)

b) 58.90 ft² (2 dp)

Step-by-step explanation:

<u>Formula</u>

\textsf{Arc length}=r \theta

\textsf{Area of a sector}=\dfrac{1}{2}r^2 \theta

\quad \textsf{(where r is the radius and}\:\theta\:{\textsf{is the angle in radians)}

<u>Calculation</u>

Given:

  • \theta=\dfrac{3 \pi}{2}
  • r = 5 ft

\begin{aligned}\implies \textsf{Arc length} & =r \theta\\& = 5\left(\dfrac{3 \pi}{2}\right)\\& = \dfrac{15}{2} \pi \\& = 23.56\: \sf ft\:(2\:dp)\end{aligned}

\begin{aligned} \implies \textsf{Area of a sector}& =\dfrac{1}{2}r^2 \theta\\\\ & = \dfrac{1}{2}(5^2) \left(\dfrac{3 \pi}{2}\right)\\\\& = \dfrac{25}{2}\left(\dfrac{3 \pi}{2}\right)\\\\ & = \dfrac{75}{4} \pi \\\\& = 58.90 \: \sf ft^2\:(2\:dp)\end{aligned}

6 0
3 years ago
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