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3 years ago

12

Month after month, Jessica is overspending in her budget and decreasing her savings account balance as a result. Which action wo

uld you recommend she take immediately
A.write a new lease so she is paying 25% less in rent each month
B.Stop paying her credit card bill and put the money into savings
C.Start packing her lunch from home every day instead of dining out
D.Quit her job so she an look for one that pays more

1 answer:

Black_prince [1.1K]3 years ago

6 0

Answer:

no

Step-by-step explanation:

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3 years ago

A right circular solid cone of base radius 42cm and height 84cm has the same volume as a solid cylinder of height 112cm find the

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Answer:

21 cm

Step-by-step explanation:

Volume of cone is given by $1/3(\pi r^2h)$

volume of cylinder is given by $\pi r^2h$

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_________________________________

For cone

radius = 42cm

height = 84

volume = $1/3(\pi *42^2*84)$ (1)

________________________________

For cone

radius = to find let it be r

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_______________________________________

since , volume of both solid are same

thus, equation 1 = equation 2

$1/3(\pi *42^2*84) = \pi *r^2*112\\=> r^2 = 1/3((\pi *42^2*84) / 112*\pi \\=> r^2 = 1/3(42*42*84/112) = 441\\=> r = \sqrt{441} \\=> r = 21$

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3 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