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GrogVix [38]
4 years ago
11

I forgot how to do this!!

Mathematics
1 answer:
Kryger [21]4 years ago
3 0
The answer would be 6:15 since it can be divided by two and equal 3:5, i think
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Nine years ago, a woman was three times as old as her son. In eight years from now, the sum of
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she would would be about 61 years and her son about 20 years

Step-by-step explanation:

78-9=69

70-8=61

61/3=20.3

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Determine if the ordered pair is a solution to the linear equation.<br> -2x – 3y =-18
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What property is 6+0=6
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The cost of controlling emissions at a firm rises rapidly as the amount of emissions reduced increases. Here is a possible model
Dahasolnce [82]

Answer:

The level of reduction that corresponds to the lowest average cost per pound of pollutant is q=\sqrt{\frac{1300}{197}}\approx 2.57pounds of pollutant per day.

The resulting average cost is $1012 per pound.

Step-by-step explanation:

We know the cost of controlling emissions C(q) = 1,300 + 197q^2 where q is the reduction in emissions (in pounds of pollutant per day) and C is the daily cost to the firm (in dollars) of this reduction.

We need to identify the objective function. The objective function is the quantity that must be made as small as possible.

In this case it is the average cost, which is given by

\bar{C}(q)=\frac{C(q)}{q} =\frac{1,300 + 197q^2}{q} = 197q+\frac{1300}{q}

Next, we want to minimize the function \bar{C}(q)= 197q+\frac{1300}{q} for this we need to find the derivative \bar{C}(q)'

\frac{d}{dq} \bar{C}(q)= \frac{d}{dq} (197q+\frac{1300}{q})\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\\bar{C}(q)'=\frac{d}{dq}\left(197q\right)+\frac{d}{dq}\left(\frac{1300}{q}\right)\\\\\bar{C}(q)'=197-\frac{1300}{q^2}

Now, we set the derivative equal to zero and solve for q to find critical points. Critical points are where the slope of the function is zero or undefined.

197-\frac{1300}{q^2}=0\\197q^2-\frac{1300}{q^2}q^2=0\cdot \:q^2\\197q^2-1300=0\\197q^2=1300\\q^2=\frac{1300}{197}\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\q=\sqrt{\frac{1300}{197}},\:q=-\sqrt{\frac{1300}{197}}

We reject q=-\sqrt{\frac{1300}{197}} because we can have negative reduction in emissions.

We apply the Second Derivative Test,

<em>If f(x_0)>0, then f has a local minimum at x_0</em>

We find \bar{C}(q)''

\frac{d}{dq} \bar{C}(q)'=\frac{d}{dq} (197-\frac{1300}{q^2})\\\\ \bar{C}(q)''= \frac{2600}{q^3}

\bar{C}(\sqrt{\frac{1300}{197}})''= \frac{2600}{(\sqrt{\frac{1300}{197}})^3}\\\\\bar{C}(\sqrt{\frac{1300}{197}})''=\frac{2600}{\frac{10^3\cdot \:13\sqrt{13}}{197\sqrt{197}}}\\\\\bar{C}(\sqrt{\frac{1300}{197}})''=\frac{197\sqrt{2561}}{65}

We can see that \bar{C}(\sqrt{\frac{1300}{197}})''>0, then \bar{C}(q) has a local minimum at q=\sqrt{\frac{1300}{197}}.

The level of reduction that corresponds to the lowest average cost per pound of pollutant is q=\sqrt{\frac{1300}{197}}\approx 2.57pounds of pollutant per day.

\bar{C}(\sqrt{\frac{1300}{197}})=197(\sqrt{\frac{1300}{197}})+\frac{1300}{\sqrt{\frac{1300}{197}}}\\\\\bar{C}(\sqrt{\frac{1300}{197}})=10\sqrt{2561}+10\sqrt{2561}\\\\\bar{C}(\sqrt{\frac{1300}{197}})=20\sqrt{2561}\approx 1012

And the resulting average cost is $1012 per pound.

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Three sixth-grade classes raised $25.50, $49.75,
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Answer:

$37.50

Step-by-step explanation:

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