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Mumz [18]
3 years ago
14

Betsy started a business 2 years ago. She had 8 customers in the first year. She added 18

Mathematics
1 answer:
Tatiana [17]3 years ago
5 0

Answer:

Betsy isn't correct. Betsy's calculations are wrong as she made her calculations by multiplying the base number of customers by 3 and then summing it all up for the 4 years under review instead of a continuous multiplication by 3.

26 × 3 × 4 is different from 26 × 3⁴

- If Betsy meets her target for the next 4 years, she will have 2106 customers in 4 years time.

Step-by-step explanation:

- Betsy started her business 2 years ago.

- In the first year, Betsy has 8 customers.

- In the second year, Betsy added 18 customers,

- So, her customers are now 8+18 = 26

- Betsy's goal is that in each of the next 4 years, she will have three times as many customers as the year before.

If the base number of customers is truly 26 this year,

Next year, Betsy would have 26 × 3 = 78 customers.

In 2 years, she would have 78 × 3 = 234

Thus indicating that in t years from now, the number of her customers will be

N(t) = N₀ × 3ᵗ

where

N(t) = number of customers t time from now

N₀ = number of customers at the moment = 26

t = number of years from now.

N(t) = 26 × 3ᵗ

In 4 years from now, Betsy will have

N(t=4) = 26 × 3⁴ = 2,106

So, it is evident that Betsy's calculations are wrong as she made her calculations by multiplying the base number of customers by 3 and then summing it all up for the 4 years under review instead of a continuous multiplication by 3.

26 × 3 × 4 is different from 26 × 3⁴

- If Betsy meets her target for the next 4 years, she will have 2106 customers in 4 years time.

Hope this Helps!!!

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Answer:

{x = 1 , y=1, z=0

Step-by-step explanation:

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{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+3 y + 2 z = 3 | (equation 2)

{2 x + 3 y + 3 z = 5 | (equation 3)

Add equation 1 to equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

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{0 x+5 y + 6 z = 5 | (equation 3)

Swap equation 2 with equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+3 y + 2 z = 3 | (equation 3)

Subtract 3/5 × (equation 2) from equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+0 y - (8 z)/5 = 0 | (equation 3)

Multiply equation 3 by 5/8:

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{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+0 y - z = 0 | (equation 3)

Multiply equation 3 by -1:

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{0 x+5 y + 6 z = 5 | (equation 2)

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Subtract 6 × (equation 3) from equation 2:

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Divide equation 2 by 5:

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Subtract 2 × (equation 2) from equation 1:

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