A man buys a horse for $60. He sells the horse for $70. He then buys the horse back for $80. And he sells the horse again for $90. In the end, how much money did the man make or lose? Or did he break even?

Answer:

x = 25

Step-by-step explanation:

Those 2 angles together form a straight line which is equal to 180°. So you can set up an equation:

3x + 18 = 93

-18 -18

3x = 75

Divide both sides by 3

x = 25

Here we have to use the fact that the domain of the function is equal to the range of the inverse function and range of the function is equal to the domain of the inverse function .

In the given question , the domain of the given function is given. So the range of the given option is same as the domain.

Therefore the correct option is c.

The appropriate choice is ...

... D. 7 19/21

Answer:

I Do

Step-by-step explanation:

Answer and Step-by-step explanation:

Let x and y be two positive integers and their sum is 14:

X + y = 14

And the sum of square of this number is:

f = x2 + y2

 = x2+ (14 – x)2

Differentiate with respect to x, we get:

F’(x) = [ x2 + (14 – x)2]’ = 0

        2x + 2(14-x)(-1) = 0

        2x +( 28 – 2x)(-1) = 0

     2x – 28 +2x = 0

        2x + 2x = 28

         4x = 28

       X = 7

Hence, y = 14 – x = 14 -7 = 7

Now taking second derivative test:

F”(x) > 0

For x = y = 7,f reaches its maximum value:

(7)2 + (7)2 = 49 + 49

                   = 98

F at endpoints x Є [ 0, 14]

F(0) = 02 + (14 – 0)2

       =  196

F(14) = (14)2 + (14 – 14)2

  = 196

Hence the sum of squares of these numbers is minimum when x = y = 7

And maximum when numbers are 0 and 14.

Final answer:

To find two positive integers such that their sum is 14, and the sum of their squares is minimized, we need to consider all possible pairs of positive integers and calculate their sums of squares. The pair (6, 8) has the minimum sum of squares of 100. To find two positive integers such that their sum is 14, and the sum of their squares is maximized, the pairs (1, 13) and (2, 12) both have the maximum sum of squares of 170. Since we need to find two positive integers, the pair (1, 13) is the answer.

Explanation:

To find two positive integers such that their sum is 14 and the sum of their squares is minimized, we need to consider all possible pairs of positive integers that add up to 14 and calculate their sums of squares. Let's list all the pairs:

• 1 and 13: 1^2 + 13^2 = 170
• 2 and 12: 2^2 + 12^2 = 148
• 3 and 11: 3^2 + 11^2 = 130
• 4 and 10: 4^2 + 10^2 = 116
• 5 and 9: 5^2 + 9^2 = 106
• 6 and 8: 6^2 + 8^2 = 100
• 7 and 7: 7^2 + 7^2 = 98

From the list, we can see that the pair (6, 8) has the minimum sum of squares, which is 100.

Similarly, to find two positive integers such that their sum is 14 and the sum of their squares is maximized, we need to again consider all possible pairs and calculate their sums of squares. Let's list the pairs:

• 1 and 13: 1^2 + 13^2 = 170
• 2 and 12: 2^2 + 12^2 = 148
• 3 and 11: 3^2 + 11^2 = 130
• 4 and 10: 4^2 + 10^2 = 116
• 5 and 9: 5^2 + 9^2 = 106
• 6 and 8: 6^2 + 8^2 = 100
• 7 and 7: 7^2 + 7^2 = 98

From the list, we can see that the pair (1, 13) and the pair (2, 12) both have the maximum sum of squares, which is 170. Since we need to find two positive integers, the pair (1, 13) is the answer.

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Answer:

b) The 2nd Derivative test shows us the change of sign and concavity at some point. c) At which point the concavity changes or not. This is only possible with the 2nd derivative test.

Step-by-step explanation:

a) To find the critical numbers, or critical points of:

1) The procedure is to calculate the 1st derivative of this function. Notice that in this case, we'll apply the Product Rule since there is a product of two functions.

2) After that, set this an equation then find the values for x.

b) The Second Derivative Test helps us to check the sign of given critical numbers.

Rewriting f'(x) factorizing:

Applying product Rule to find the 2nd Derivative, similarly to 1st derivative:

1) Setting this to zero, as an equation:

2) Now, let's define which is the inflection point, the domain is as a polynomial function:

Looking at the graph.

Plugging these inflection points in the original equation to get y coordinate:

We have as Inflection Points and their respective y coordinates (Converting to approximate decimal numbers)

Inflection Point and Local Minimum

Inflection Point and Saddle Point

Inflection Point Local Maximum

(Check the graph)

c) At which point the concavity changes or not. This is only possible with the 2nd derivative test.

At

Local Minimum

(Saddle Point)

Final answer:

To find the critical numbers of the function f(x) = x^6(x - 2)^5, we need to set the first derivative equal to zero and solve for x. The Second Derivative Test tells us the behavior of the function at the critical numbers, while the First Derivative Test tells us the behavior of the function based on the sign change of the derivative at the critical numbers.

Explanation:

The critical numbers of the function f(x) = x^6(x - 2)^5 can be found by taking the first and second derivatives of the function. The first derivative is f'(x) = 6x^5(x - 2)^5 + 5x^6(x - 2)^4 and the second derivative is f''(x) = 30x^4(x - 2)^5 + 20x^5(x - 2)^4.

To find the critical numbers, we need to set the first derivative equal to zero and solve for x: 6x^5(x - 2)^5 + 5x^6(x - 2)^4 = 0. We can solve this equation using factoring or by using the Zero Product Property. Once we find the values of x that make the first derivative zero, we can evaluate the second derivative at those values to determine the behavior of the function at those critical numbers.

The Second Derivative Test tells us that if the second derivative is positive at a critical number, then the function has a local minimum at that point. If the second derivative is negative at a critical number, then the function has a local maximum at that point. If the second derivative is zero, the test is inconclusive and we need to use additional information to determine the behavior of the function. The First Derivative Test tells us that if the derivative changes sign from negative to positive at a critical number, then the function has a local minimum at that point. If the derivative changes sign from positive to negative at a critical number, then the function has a local maximum at that point.

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