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Maslowich
2 years ago
5

NEED HELP!!!!!!!!!!!

Mathematics
1 answer:
lina2011 [118]2 years ago
4 0

Answer:

6

Step-by-step explanation:

do you want an explanation?

btw, plz brainliest :)

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Vladimir [108]
So you start at 0 then (1,3) , (2,12) and the other side would be (-1,3) , (-2,12)
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A snack food manufacturer buys corn for tortilla chips from two cooperatives, one in Iowa and one in Illinois. The price per uni
aleksandrvk [35]

Answer:

a. <u>x1 = No. of units to purchase from Iowa</u>

   <u>x2 = No. of units to purchase from Illinois</u>

<u></u>

b. <u>Min 6x1 + 5.5x2</u>

<u></u>

c. <u>x1 + x2 ≥ 12000</u>

   <u>x1 ≤ 8000</u>

   <u>x2 ≥ 6000</u>

Step-by-step explanation:

a. We can consider the variables x1 and x2 as:

<u>x1 = No. of units to purchase from Iowa</u>

<u>x2 = No. of units to purchase from Illinois</u>

<u></u>

b. Price per unit of Iowa corn = $6

   Price per unit of Illinois corn = $5.5

Objective function that would minimize the total cost can be written as:

<u>Min 6x1 + 5.5x2</u>

<u></u>

c. The manufacturer needs at least 12000 units of corn which means that the combined number of units from Iowa and Illinois must be greater than or equal to  12000. So, we can write:

x1 + x2 ≥ 12000

The Iowa cooperative can supply up to 8000 units which means that the value of x1 must not be greater than 8000. So, we can write:

x1 ≤ 8000

Similarly, the Illinois cooperative can supply at least 6000 units which means that the value of x2 must not be less than 6000. So, we can write:

x2 ≥ 6000.

The constraints for these conditions are:

<u>x1 + x2 ≥ 12000</u>

<u>x1 ≤ 8000</u>

<u>x2 ≥ 6000</u>

6 0
2 years ago
Determine what type of model best fits the given situation:
lyudmila [28]

Let value intially be = P

Then it is decreased by 20 %.

So 20% of P = \frac{20}{100} \times P = 0.2P

So after 1 year value is decreased by 0.2P

so value after 1 year will be = P - 0.2P (as its decreased so we will subtract 0.2P from original value P) = 0.8P-------------------------------------(1)

Similarly for 2nd year, this value 0.8P will again be decreased by 20 %

so 20% of 0.8P = \frac{20}{100} \times 0.8P = (0.2)(0.8P)

So after 2 years value is decreased by (0.2)(0.8P)

so value after 2 years will be = 0.8P - 0.2(0.8P)

taking 0.8P common out we get 0.8P(1-0.2)

= 0.8P(0.8)

=P(0.8)^{2}-------------------------(2)

Similarly after 3 years, this value P(0.8)^{2} will again be decreased by 20 %

so 20% of P(0.8)^{2}  \frac{20}{100} \times P(0.8)^{2} = (0.2)P(0.8)^{2}

So after 3 years value is decreased by (0.2)P(0.8)^{2}

so value after 3 years will be = P(0.8)^{2}   - (0.2)P(0.8)^{2}

taking P(0.8)^{2} common out we get P(0.8)^{2}(1-0.2)

P(0.8)^{2}(0.8)

P(0.8)^{3}-----------------------(3)

so from (1), (2), (3) we can see the following pattern

value after 1 year is P(0.8) or P(0.8)^{1}

value after 2 years is P(0.8)^{2}

value after 3 years is P(0.8)^{3}

so value after x years will be P(0.8)^{x} ( whatever is the year, that is raised to power on 0.8)

So function is best described by exponential model

y = P(0.8)^{x} where y is the value after x years

so thats the final answer

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