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DiKsa [7]
3 years ago
12

During 2005, a company produced an average of 2,000 products per month. How many products will the company need to produce from

2006 through 2008 in order to increase its monthly average for the period from 2005 through 2008 by 200% over its 2005 average?
Mathematics
1 answer:
Svet_ta [14]3 years ago
3 0

Answer:

The company needs to produce 264,000 units from 2006 through 2008.

Step-by-step explanation:

Let x represent units produced by company from 2006 to 2008.

We have been given that during 2005, a company produced an average of 2,000 products per month.

The total units produced by company in 2005 would be 12\times 2,000=24,000 units.

The total units produced by company from 2005 to 2008 would be x+24,000 units.

The average of units produced during four years would be total units (x+24,000) divided by total years (4) that is \frac{x+24,000}{4}

We are told that the average of the 4 years should be 200% more than in 2005. Let us find average for 4 years as:

24,000+24,000\times \frac{200}{100}\\\Rightarrow 24,000+24,000\times 2\\=24,000+48,000\\=72,000

We can represent all this information in an equation as:

\frac{x+24,000}{4}=72,000

\frac{x+24,000}{4}*4=72,000*4

x+24,000=288,000

x+24,000-24,000=288,000-24,000

x=264,000

Therefore, the company needs to produce 264,000 units from 2006 through 2008.

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Sin(xy)-x=0 find dy/Dx
MariettaO [177]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2989024

——————————

You have  y  as an implicit function of  x:

sin(xy) – x = 0


Use implicit differentiation. As  y  is a function of  x, then you must apply the chain rule there:

  d                                 d
—— [ sin(xy) – x ]  =  —— (0)
 dx                               dx


  d                           d               d
—— [ sin(xy) ]  –  —— (x)  =  —— (0)
 dx                         dx             dx

  d
—— [ sin(xy) ]  –  1  =  0
 dx

  d
—— [ sin(xy) ]  =  1
 dx
 
                   d
cos(xy)  ·  —— (xy) = 1
                  dx


Now, apply the product rule for that last derivative:

\mathsf{cos(xy)\cdot \left[\dfrac{d}{dx}(x)\cdot y+x\cdot \dfrac{dy}{dx}\right]=1}\\\\\\
\mathsf{cos(xy)\cdot \left[1\cdot y+x\cdot \dfrac{dy}{dx}\right]=1}\\\\\\
\mathsf{y\,cos(xy)+x\,cos(xy)\cdot \dfrac{dy}{dx}=1}


              dy
Isolate  —— :
              dx

                   dy
x cos(xy) · ——  =  1 – y cos(xy)
                   dx


Assuming  x cos(xy) ≠ 0,

 dy          1 – y cos(xy)
——  =  ————————    <———    this is the answer.
 dx            x cos (xy)


I hope this helps. =)

4 0
3 years ago
Solve for y: 3d = 7y + 5
pogonyaev

Answer:

d = \frac{7y+5}{3}

Step by step explanation:

3d = 7y+5

<em>Divide both sides by 3</em>

<em />\frac{3d}{3}<em>= </em>\frac{7y}{3}+ \frac{5}{3}

<em>Simplify</em>

\frac{7y+5}{3}

7 0
2 years ago
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