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noname [10]
3 years ago
8

A granola bite contains 27 calories. Most of the calories come from c grams of carbohydrates. The rest come from other ingredien

ts. One gram of carbohydrate contains 4 calories. The equation 4c + 5 = 27 represents the relationship between these quantities. Step 1: Answer the following questions: a. What could the 5 represent in this situation? b. Priya said that neither 8 nor 3 could be the solution to the equation. Explain why she is correct. c. Find the solution to the equation.
Mathematics
1 answer:
joja [24]3 years ago
8 0

Answer:

a. 5 represents the number of calories that come from other ingredients.

b. Neither 8 nor 3 satisfy the given equation, so neither 8 nor 3 can be the solution to the equation.

c. <em>c = 5.5</em>

Step-by-step explanation:

We are given the equation:

4c + 5 = 27

Where c represents the grams of carbohydrates

27 is the total number of calories that are received by the granola bite.

4 is the number of calories from 1 gram of carbohydrates and

5 is the number of calories that are received from other ingredients.

Answer of Part a:

5 is the number of calories that are received from other ingredients.

Part b:

Let us take left hand side (LHS) and put values of c = 8 and 3 one by one.

So, LHS becomes:

4 \times 8 + 5 = 37 \neq RHS\ i.e.\ 27

4 \times 3 + 5 = 17 \neq RHS\ i.e.\ 27

So, Priya is correct that neither 8 or 3 could be solution.

Part c:

4c + 5 = 27\\\Rightarrow 4c = 22\\\Rightarrow \bold{c = 5.5}

So, solution to the equation is: <em>c = 5.5</em>

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The answer is shown below

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\frac{dy}{dt}=ky(1-y)\\\frac{dy}{y(1-y)}=kdt\\\int\limits {\frac{dy}{y(1-y)}} \, =\int\limit {kdt}\\\int\limits {\frac{dy}{y}} +\int\limits {\frac{dy}{1-y}}  =\int\limit {kdt}\\\\ln(y)-ln(1-y)=kt+c\\ln(\frac{y}{1-y}) =kt+c\\taking \ exponential \ of\ both \ sides\\\frac{y}{1-y} =e^{kt+c}\\\frac{y}{1-y} =e^{kt}e^c\\let\ A=e^c\\\frac{y}{1-y} =Ae^{kt}\\y=(1-y)Ae^{kt}\\y=\frac{Ae^{kt}}{1+Ae^{kt}} \\at \ t=0,y=10\%\\0.1=\frac{Ae^{k*0}}{1+Ae^{k*0}} \\0.1=\frac{A}{1+A} \\A=\frac{1}{9} \\

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