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andrezito [222]
3 years ago
7

All info in the pic. Please help. Thank you!

Mathematics
1 answer:
adelina 88 [10]3 years ago
6 0
You can write an equation.
KL+LM=KM
Enter the info
x+.5+3x-2=3x+1.5
Combine like terms
4x-1.5=3x+1.5
Add 1.5 to both sides
4x-1.5+1.5=3x+1.5+1.5
4x=3x+3
Subtract 3x from both sides
4x-3x=3x+3-3x
x=3

Then, enter 3 into the expression to find LM
3x-2
3(3)-2
9-2
7 
LM is worth 7.
I hope this helped!

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WILL GIVE BRAINIEST. Tony drives 18 miles to pick up his friend at his house. Then he drives at a constant speed of 40 miles per
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Answer:

A. C.

Step-by-step explanation:

A is right because you would use y as miles and x as time multiplied by 40 the constant and then plus 18 for the previous miles. C is right because 18 is the initial value with 18 being added

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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 2.57pounds 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.

5 0
3 years ago
Lashonda is saving money to buy a game. So far she has saved $6, which is two-thirds of the total cost of the game. How much doe
aliina [53]

Answer:

The game costs $9

Step-by-step explanation:

I wrote an equation using the variable "t" for the total cost of the game.

6=2/3t

t=6/2/3

t=(6/1)(3/2)

t=18/2

t=9

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3 years ago
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