All categories

4 years ago

How many triangles can be made if one angle is 95° and another angle is acute?

1 answer:

8 0

Theoretically, the answer will be infinite.

Since one angle of 95 degrees means the other two angles must be acute. You can pick any two real numbers with the sum of 85.

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A recipe for a smoothie calls for 0.75 pound of strawberries. If Kerry has 31.5 pounds of strawberries, how many batches of the

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42 batches. 31.5 divided by .75 = 42.

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3 years ago

Please help me on this surface area question

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The answer for this question is letter c

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3 years ago

Find the slope of the line that passes through the points (4, 4) and (5,3).

Alex777 [14]

Answer:

the slope of the line is -1

Step-by-step explanation:

slope= change in y/change in x

slope=(3-4)/(5-4)

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3 years ago

Put the following numbers in order from smallest to largest.

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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.

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

And the resulting average cost is $1012 per pound.

5 0

3 years ago