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3 years ago

Math, Stumped me once more

Mathematics

1 answer:

EastWind [94]3 years ago

4 0

Answer:

$B.Alternate\ Exterior\ Angles$

Step-by-step explanation:

$As\ the\ exterior\ angles\ lie\ on\ the\ opposite\ sides\ of\ the\ transversal,\\that\ cuts\ two\ parallel\ lines,\ the\ set\ of\ 'Alternate\ Exterior\ Angles'\ formed\\are\ equal.$

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Can you answer it pls and by the way i am in 5th grade

timofeeve [1]

Step-by-step explanation:

3×3 = 9, so it's greater than 3

3×4/3 = 4, so it's greater than 3

3×5/5 = 3, so it's equal to 3

3x4/9 = 4/3, so it's less than 3

3x5/6 = 5/2, so it's less than 3

4 0

2 years ago

QUICK need help on this

lubasha [3.4K]

It’s 2 because if you finish the equation

8 0

4 years ago

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 $q=\sqrt{\frac{1300}{197}}\approx 2.57$ pounds 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 $q=\sqrt{\frac{1300}{197}}\approx 2.57$ pounds of pollutant per day.

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