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3 years ago

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A shipment of 288 reams of paper was delivered. Each of the 30 classrooms recieved an equal share of equipment paper. Any extra

reams of paper were stored. After the paper was distributed to the classrooms, how many reams of paper were stored?

1 answer:

Andrei [34K]3 years ago

8 0

288 divided by 30 equals 9 with a remainder of 18

= 9 R 18
<span>= 9 18/30</span>

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Bob's monthly bill is made up of a $10 fee plus $0.05 per minute. Write an expression that represents Bob's monthly phone if he

Salsk061 [2.6K]

Answer:

10 + 0.05m; m = 350 minutes

Bob's phone bill will be $27.50

Step-by-step explanation:

$10 + ($0.05 x 350 minutes) = $27.50

6 0

3 years ago

Solve this: (a^3-2a^2)-(3a^2-4a^3)

Jobisdone [24]

(a^3-2a^2)-(3a^2-4a^3)
(a^3-2a^2)-3a^2+4a^3
5a^3-5a^2

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3 years ago

Brainliest plus 10 points! simplify<br> 6y^2-6/8y^2+8y÷3y-3/4y^2+4

lesya692 [45]

Answer:

(y² +1)/y

Step-by-step explanation:

Invert the denominator fraction and multiply. Factor the difference of squares.

$\displaystyle\frac{\left(\frac{6y^2-6}{8y^2+8y}\right)}{\left(\frac{3y-3}{4y^2+4}\right)}=\frac{6(y^2-1)}{8y(y+1)}\cdot\frac{4(y^2+1)}{3(y-1)}\\\\=\frac{24(y+1)(y-1)(y^2+1)}{24y(y+1)(y-1)}=\frac{y^2+1}{y}$

6 0

3 years ago

I can't figure out how to solve this

Alex

Because both lines AT and AN are tangents, they are equal to each other.

Because angle ATN is 70 degrees, then angle ANT is also 70 degrees

The 3 inside angles of a triangle = 180 degrees,
so angle A would = 180 - 70 - 70
Angle A = 180 -70-70 = 40 degrees

7 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.

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

And the resulting average cost is $1012 per pound.

5 0

3 years ago