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stich3 [128]
4 years ago
15

Which statement correctly explains the chart?

Mathematics
2 answers:
makvit [3.9K]4 years ago
7 0

The chart shows a production possibilities schedule for Sabrina’s Soccer.

Combination: Soccer balls: Soccer nets:

A 10 0

B 8 1

C 6 2

D 4 3

E 2 4

F 0 5

Which statement correctly explains the chart?

A. The opportunity cost of producing one soccer net is eight soccer balls.

B. The opportunity cost of producing two soccer nets is two soccer balls.

C. The opportunity cost of producing two soccer balls is one soccer net.

D. The opportunity cost of producing four soccer balls is three soccer nets.

The opportunity cost of producing two soccer balls is one soccer net.

Answer: Option 3

<u>Explanation:</u>

Opportunity cost is when a particular option is chosen from the alternatives given, the opportunity cost is the cost incurred by not enjoying the benefit associated with the best alternative choice.

The problem of the opportunity cost occurs because the resources given in the economy are limited in availability and there fore because of that there has to be some choices that are to be made among the alternatives given in the economy.

In this example it shows that for producing two soccer balls, the opportunity cost is one soccer net.

vovikov84 [41]4 years ago
5 0

Answer:

C. The opportunity cost of producing two soccer balls is one soccer net.

Step-by-step explanation:

EDg

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If x2+1/x2=18, find the value of x3−1/x3
Elden [556K]

Answer:

There are 2 answers : 76 or -76

Step-by-step explanation:

x^2+\dfrac{1}{x^2} =18\\\\(x-\dfrac{1}{x} )^2=x^2+\dfrac{1}{x^2} -2=18-2=16\\\\x-\dfrac{1}{x}=4\ or\ x-\dfrac{1}{x}-4\\\\(x-\dfrac{1}{x} )^3=x^3-\dfrac{1}{x^3} -3x^2*\dfrac{1}{x}+3x*\dfrac{1}{x^2}\\\\=x^3-\dfrac{1}{x^3}-3(x-\dfrac{1}{x})\\\\x^3-\dfrac{1}{x^3}=(x-\dfrac{1}{x} )^3+3(x-\dfrac{1}{x})\\\\if\ x-\dfrac{1}{x}=4\ then \ x^3-\dfrac{1}{x^3}=4^3+3*4=64-12=76\\\\if\ x-\dfrac{1}{x}=-4\ then \ x^3-\dfrac{1}{x^3}=(-4)^3+3*(-4)=-64-12=-76\\\\

7 0
3 years ago
Read 2 more answers
 For the equation 2n - 3 = 7
Stells [14]

Answer:

The variable is n

The coefficient is 2

And the solution to n is 10.

Step-by-step explanation:

The number in the equation is the variable. The coefficient is the number before the variable. And a number next to a letter means you multiply. So 2*5=10 10-3=7

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3 years ago
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(2, 50, 50, 51, 51, 52, 53, 54, 54, 55, 55, 55, 56, 56, 57} What is the best measure of the central tendency of the data? mean m
Alexxx [7]
The median is----54
The mean is----Approximately 50
The mode(s) is/are----50, 51, 54, 55, and 56
The range is---- Approximately 29 if you round
7 0
3 years ago
Use the definition of continuity to determine whether f is continuous at a.
dmitriy555 [2]
f(x) will be continuous at x=a=7 if
(i) \displaystyle\lim_{x\to7}f(x) exists,
(ii) f(7) exists, and
(iii) \displaystyle\lim_{x\to7}f(x)=f(7).

The second condition is immediate, since f(7)=8918 has a finite value. The other two conditions can be established by proving that the limit of the function as x\to7 is indeed the value of f(7). That is, we must prove that for any \varepsilon>0, we can find \delta>0 such that

|x-7|

Now,


|f(x)-f(7)|=|5x^4-9x^3+x-8925|

Notice that when x=7, we have 5x^4-9x^3+x-8925=0. By the polynomial remainder theorem, we know that x-7 is then a factor of this polynomial. Indeed, we can write

|5x^4-9x^3+x-8925|=|(x-7)(5x^3+26x^2+182x+1275)|=|x-7||5x^3+26x^2+182x+1275|

This is the quantity that we do not want exceeding \varepsilon. Suppose we focus our attention on small values \delta. For instance, say we restrict \delta to be no larger than 1, i.e. \delta\le1. Under this condition, we have

|x-7|

Now, by the triangle inequality,


|5x^3+26x^2+182x+1275|\le|5x^3|+|26x^2|+|182x|+|1275|=5|x|^3+26|x|^2+182|x|+1275

If |x|, then this quantity is moreover bounded such that

|5x^3+26x^2+182x+1275|\le5\cdot8^3+26\cdot8^2+182\cdot8+1275=6955

To recap, fixing \delta\le1 would force |x|, which makes


|x-7||5x^3+26x^2+182x+1275|

and we want this quantity to be smaller than \varepsilon, so


6955|x-7|

which suggests that we could set \delta=\dfrac{\varepsilon}{6955}. But if \varepsilon is given such that the above inequality fails for \delta=\dfrac{\varepsilon}{6955}, then we can always fall back on \delta=1, for which we know the inequality will hold. Therefore, we should ultimately choose the smaller of the two, i.e. set \delta=\min\left\{1,\dfrac{\varepsilon}{6955}\right\}.

You would just need to formalize this proof to complete it, but you have all the groundwork laid out above. At any rate, you would end up proving the limit above, and ultimately establish that f(x) is indeed continuous at x=7.
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Could a right triangle ever be an equilateral triangle?
andriy [413]

Answer: An equilateral triangle can NOT be a right triangle.

Step-by-step explanation:

In a right triangle, however, the 3 angles can NOT be congruent. This is because in a right triangle, one angle equals 90 degrees. Therefore, you can't have 3 angles equal 90 degrees.

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3 years ago
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