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Laura P.

Tutor with 3 years experience

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Biology

TutorMe

Question:

Explain the way that amylose is chemically digested, absorbed and stored in the liver.

Laura P.

Answer:

Amylose digestion begins in the mouth. In the saliva, the enzyme amylase. Amylase will digest the amylose into glucose. The enzyme becomes inactive in the stomach, but it is secreted again by the pancreas into the small intestine. In the small intestine, the glucose is absorbed into the enterocytes. On the apical side, this is done by symporters that absorb sodium along their concentration gradient in order to provide the energy to absorb the glucose against their gradient. In order to maintain the sodium gradient, sodium potassium pumps on the basal side moves the sodium out of the enterocyte. The glucose is transported into the bloodstream using a glucose uniporter along the concentration gradient. In the liver, the glucose is converted into glycogen for storage in the liver when insulin is present.

Calculus

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Question:

You work for a company that is desk chairs. However, your company is unsure as to how to price it. You also know that you will make no money if it costs less than $0.50. From your research, you also know that when they cost $5, 1,000 people will buy it. You also know that for each $0.25 price increase 20 fewer people will buy it. Your goal is to determine the best price to sell them at and how much money that will make the company.

Laura P.

Answer:

From the information that the question is giving us, we know that the dependent variable will need to be the amount of money that the company is making. Next, we know that price can be calculated by multiplying the number of units that are sold by the price. From this, we can define the following variables: Let C be the total amount of money made by the company. Let p be the price of the units. Let n be the number of units sold, or the number of people that are buying it. We can write an equation as: $$C = pn$$ The issue we have now is that there are too many variables. However, we know some information that is given in the question. Since we know that when the price is $0.5, there will be no profits, this means that this will be the cost of the chocolates to make by the company. Therefore, this must be subtracted from the price. The equation will be as follows: $$C = n(p-0.5)$$ Now, we can use the other two pieces of information to complete the question as follows: $$C=(1000-20n)(0.25n-0.5)$$ Now that we have two variables, we can optimize. In order to do this, we will differentiate the function. However, to make this the easiest, we will first expand the quadratic. $$C=250n-500-5n^{2}+10n$$ $$C=-5n^{2}+260n-500$$ Next, we can differentiate: $$C'=-10n+260$$ We know that the highest level of profit will be when $$C' = 0$$ because it is a quadratic. $$0=-10n+260$$ $$-260 =-10n$$ $$n=26$$ From the question, we need to determine the maximum profits. To determine the maximum profits, we use the equation that we differentiated and substitute in the number of sales. $$C=-5(26)^{2}+260(26)-500$$ $$C=2460$$ To determine the optimal price, we can use the original equation and substitute in the number of units sold and cost to solve for the price $$2460=26(p-0.5)$$ $$2460=26p-13$$ $$2473=26p$$ $$95.12=p$$

Algebra

TutorMe

Question:

Determine the value of x in: $$\frac{x-1}{x+2} = 2x$$

Laura P.

Answer:

First, you will want to get rid of the fraction so you can multiply both sides by $$x+2$$: $$\frac{(x-1)(x+2)}{x+2}$$ = $$(2x)(x+2)$$ On the left side of the equation, this will cancel out the $$x+2$$ in the numerator and the denominator, leaving us with: $$x-1$$ = $$(2x)(x+2)$$ Next, we should expand the brackets on the right side of the equation. When we do this, we get: $$x-1$$ = $$2x^{2}+4x$$ Now, we have a quadratic. This means that to solve for $$x$$, we will need to get 0 on the ones of the equation with everything else on the other side. To do this, we can subtract $$x-1$$ from both sides to get: $$0$$ = $$2x^{2}+4x-(x-1)$$ Next, we will need to expand the brackets on the left side. $$0$$ = $$2x^{2}+4x-x+1$$ Now we can combine like terms: $$0$$ = $$2x^{2}+3x+1$$ This is where we need to use our knowledge of quadratics. To solve for the possible values of $$x$$, we factor: $$0$$ = $$(2x+1)(x+1)$$ Since we know that when we multiply anything by $$0$$, we can $$0$$, only one of the terms needs to equal $$0$$. Therefore, we can independently set each one equal to $$0$$. Let's start with the one closest to the equals sign and solve for x. $$0$$ = $$(2x+1)$$ $$-1$$ = $$2x$$ $$\frac{-1}{2}$$ = $$x$$ For the other one: $$0$$ = $$(x+1)$$ $$-1$$ = $$x$$ Therefore, the two possible values of x are $$-1$$ and $$\frac{-1}{2}$$

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