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ICE Princess25 [194]
1 year ago
9

Can someone help me with this?

Mathematics
1 answer:
erik [133]1 year ago
7 0

The equilibrium price is $12.

<h3>What is the equilibrium price ?</h3>

Equilibrium is the point where the quantity demanded is equal to the quantity supplied. The price at equilibrium is known as the equilibrium price and the quantity at equilibrium is known as the equilibrium quantity.

When shown on a graph, equilibrium is the point where the quantity demanded curve is equal to the quantity supplied curve.

When there is equilibrium, the equation of quantity demanded would be equal to the equation of quantity supplied.

-280 + 40p = 800  - 50p

In order to determine the value of p, take the following steps:

Combine similar terms: 800 + 280 = 40p + 50p

Add similar terms = 1080 = 90p

Divide both sides of the equation by 90 : 1080 / 90 = 12

To learn more about equilibrium, please check: brainly.com/question/26075805

#SPJ1

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11Alexandr11 [23.1K]

To factor: Notice that when you factor it should turn out to be a binomial x binomial.

x^2-10x+25

When factoring you can look at the operation at the end right before the last digit.  If the operation is positive then you will use the first operation in the binomial for factored number.  For example the ending operation is positive and the first operation is subtraction so both binomial will be subtraction.

I know it would be factored (x-5)(x-5) and I can prove it by multiplying it out.

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You have two circles, one with radius r and the other with radius R. You wish for the difference in the areas of these two circl
gtnhenbr [62]

Answer:

maximum difference of radii =(r-R)=\frac{1}{2\pi ^{2}}

Step-by-step explanation:

We know that area of circle is given by

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For circle with radius 'r' we have

A_{1}=\pi \times (r)^{2}

For circle with radius 'R' we have

A_{2}=\pi \times (R)^{2}

Now according to given condition we have

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\Rightarrow \pi r^{2}-\pi R^{2}\leq \frac{5}{\pi }\\\\\Rightarrow (r^{2}-R^{2})\leq \frac{5}{\pi ^{2}}\\\\(r+R)(r-R)\leq \frac{5}{\pi ^{2}}\\\\\because (a^{2}-b^{2})=(a+b)(a-b)\\\\(r+R)=10(Given)\\\\\Rightarrow(r-R)\leq \frac{5}{10\pi ^{2}}\\\\\therefore (r-R)\leq\frac{1}{2\pi ^{2}}

Thus maximum difference of radii =(r-R)=\frac{1}{2\pi ^{2}}

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3 years ago
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Read 2 more answers
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