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Amanda [17]
3 years ago
7

An apple pie uses 4 cups of apples and 3 cups of flour. An apple cobbler uses 2 cups of apples and 3 cups of flour. You have 16

cups of apples an 15 cups of flour. When you sell these at the Farmers market you make $3.00 profit per apple pie and $2.00 profit per apple cobbler.
Use linear programming to determine how many apple pies and how many apple cobblers you should make to maximize profit.


Let x=The number of apple pies you make. Let y=The number of apple cobblers you make Write an inequality to show the constraint on the amount of apples you have.


1a. Apple pie uses 4 cups and 3 cups of flour.


Apple cobbler uses 2 cups of apples and 3 cups of flour


You have 16 cups of apples and 15 cups of flour


Apple pie= $3.00, $2.00 apple cobbler.


X=apple pie, Y=apple cobbler


Write an inequality to show the constraint on the amount of apples you have.


4x+2y<=16


1b. Wr
Mathematics
1 answer:
Alex Ar [27]3 years ago
3 0
4x+2y<=16
3x+3y<=15

2x+y<=8
x+y<=5
3x+2y->maximize profit

X   Y    profit
1   4     11
2   3     12
3   2     13
4   0     12
3   1     11
3   2     13
2   3     12
1   4     11
0   5      10

Farmer need to make 3 Apple pies and 2 apple cobblers
Farmer will use 16 cups of apples and 15 cups of flour
profit of farmer will be $13 
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An expression for this timet-half will be given as ln(2) / λ ≈ 0.693 / λ

<h3>What is an expression?</h3>

Expression in maths is defined as the collection of the numbers variables and functions by using signs like addition, subtraction, multiplication and division.

The complete question is:

Suppose a radioactive sample initially contains N0unstable nuclei. These nuclei will decay into stable nuclei, and as they do, the number of unstable nuclei that remain, N(t), will decrease with time. Although there is no way for us to predict exactly when any one nucleus will decay, we can write down an expression for the total number of unstable nuclei that remain after a time t:

N(t)=No e−λt,

where λ is known as the decay constant. Note that at t=0, N(t)=No, the original number of unstable nuclei. N(t) decreases exponentially with time, and as t approaches infinity, the number of unstable nuclei that remain approaches zero.

Part (A) Since at t=0, N(t)=No, and at t=∞, N(t)=0, there must be some time between zero and infinity at which exactly half of the original number of nuclei remain. Find an expression for this time, t half.

Express your answer in terms of N0 and/or λ.

1) Equation given:

← I used α instead of λ just for editing facility..

Where No is the initial number of nuclei.

2) Half of the initial number of nuclei: N (t-half) =  No / 2

So, replace in the given equation:

N9t) = Ne^{-\lambda t}

3) Solving for α (remember α is λ)

N_{t-half}=\dfrac{N_o}{2}=N_oe^{-\alpha t}

\dfrac{1}{2} = e^{-\alpha t}2=e^{-alpha t}\alpha t=In(2)

αt ≈ 0.693

⇒ t = ln (2) / α ≈ 0.693 / α ← final answer when you change α for λ

Therefore expression for this timet-half will be given as ln(2) / λ ≈ 0.693 / λ

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brainly.com/question/723406

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