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sergeinik [125]
3 years ago
12

S Explain how knowing 50 times 4 =200 helps you find 500 times 400.

Mathematics
1 answer:
Lemur [1.5K]3 years ago
3 0
This is an easy way to solve multiplication with alot of zero's
Add only the numbers, then add the sum of the zeros
So with this problem we do this:-
5*4 is 20, plus the zero in the 50, so the answer is 200
Now we can do the other one:-

5*4 is 20, plus the 4 zeros that we removed is 200,000

Hope I helped. :)
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Ahat [919]

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A) 250

B) 1250

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7 0
2 years ago
A company manufactures and sells x television sets per month. The monthly cost and price-demand equations are
skelet666 [1.2K]

A) The maximum revenue is 450000$

B) The maximum profit is 216000$ when 2400 sets are manufactured and sold for 180$ each

C) When each set is taxed at ​$55​, the maximum profit is 99125$ when 1850 sets are manufactured and sold for 207.5$ each.

A)p(x)=300−(x/20​),

revenue R(x)=p*x

revenue R(x)=300x -(x2/20)

for maximum revenue dR/dx =0 ,

=>300-(2x/20)=0

=>x/10=300

=>x=3000

maximum revenue = R(3000)=300*3000 -(30002/20)

maximum revenue = R(3000)=450000$

B) profit =revenue -cost

profit P(x)=300x -(x2/20)-72000-60x

profit P(x)=240x -(x2/20)-72000

for maximum cost dP/dx =0

240 -(2x/20)=0

x=240*10

x=2400

p(2400)=300−(2400/20​)=180

profit P(2400)=240*2400 -(24002/20)-72000 =216000

The maximum profit is 216000$ when 2400 sets are manufactured and sold for 180$ each

c)

profit =revenue -cost -tax

profit P(x)=300x -(x2/20)-72000-60x-55x

profit P(x)=185x -(x2/20)-72000

for maximum cost dP/dx =0

185-(2x/20)=0

x=185*10

x=1850

p(1850)=300−(1850/20​)=207.5

profit P(1850)=185*1850 -(18502/20)-72000

profit P(1850)=99125$

When each set is taxed at ​$55​, the maximum profit is 99125$ when 1850 sets are manufactured and sold for 207.5$ each.

To know more about maximum profit check the below link:

brainly.com/question/4166660

#SPJ4

5 0
9 months ago
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