Eric leans a ladder against the roof of his house so that the ladder forms a 2004-04-02-01-00_files/i0260000.jpg angle with the ground. The roof of the house is 13 feet above the ground. How far is the bottom of the ladder from the base of the house? Round the answer to the nearest tenth.

Final answer:

To show that the given vector field is not a gradient vector field, we need to compute its curl. The curl is computed by taking the cross product of the gradient operator with the vector field. By computing the partial derivatives and simplifying, we find that the curl of the given vector field is not zero, which indicates that it is not a gradient vector field.

Explanation:

To show that the vector field is not a gradient vector field, we need to compute its curl. The given vector field is f(x,y,z) = <ycos(-2x), -2xsin(y), 0>. The curl of a vector field is defined as the cross product of the gradient operator with the vector field. In this case, the curl of f(x,y,z) is computed as:

curl(f) = ∇ × f = (∂f_z/∂y - ∂f_y/∂z)i + (∂f_x/∂z - ∂f_z/∂x)j + (∂f_y/∂x - ∂f_x/∂y)k

By computing the partial derivatives and simplifying, we find that:

Substituting these values back into the curl formula, we get:

curl(f) = (2ycos(-2x))i + 0j + (-2sin(y))k

This shows that the curl of the vector field is not zero, which means the vector field is not a gradient vector field.

Answer:

-5,5

Step-by-step explanation:

Given the derivative f '(x) ≥ 2, the function f(x) increases by at least 2 units for every increase in 'x' from 3 to 8. Therefore, the smallest possible value of f(8) is reached when the increase is exactly 2 per 'x', giving us a result of 21.

The question is about calculus, specifically related to the concept of derivatives and their role in function interval analysis. Given the information that f(3) = 11 and f '(x) ≥ 2 for 3 ≤ x ≤ 8, we are asked to find the smallest possible value of f(8).

The derivative, f '(x), represents the rate of change of the function f(x). It is mentioned that f '(x) ≥ 2, which means that the function f(x) is increasing at a rate not less than 2 units per increase in 'x' from 3 to 8.

This suggests that the smallest possible value of f(8) could be achieved by assuming that f(x) increases exactly by 2 units for every increase in 'x'. So, from x = 3 to x = 8 is a change of 5 units in 'x'. Multiply this by the rate of change 2 to obtain the total change in f(x), which is 5 * 2 = 10. Added to the initial value f(3) = 11, we get the smallest possible value for f(8) is 11 + 10 = 21.

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Final answer:

To convert 300mm to meters, divide 300 by 1000 because there are 1000 mm in a meter, resulting in 0.3 meters.

Explanation:

To convert millimeters to meters, we use the conversion factor that 1000 mm equals 1 m. Therefore, to convert 300mm to meters, you divide 300 mm by 1000 because there are 1000 mm in a single meter. So, let's carry out the calculation:

300 mm ÷ 1000 = 0.3 m

Thus, 300 mm is equal to 0.3 meters.

The output of the function (f ^ g)(30) in this context refers to first calculating the output of the g function given x = 30, then substituting this output into the f function. With g(30) = sqrt(x - 5) and f(x) = 2x - 8, we find that (f ^ g)(30) = 2.

To find (f ^ g)(30), we first need to find the output of the g function when x = 30 and then substitute this output into the f function. This is also known as the composition of functions.

First, let's find the output of the g function when x = 30. Given that g(x) = sqrt(x - 5), for x = 30, g(30) = sqrt(30 - 5) = sqrt(25) = 5.

Next, we substitute this output into the f function. Given that f(x) = 2x - 8, when x = 5, f(5) = 2(5) - 8 = 10 - 8 = 2.

Therefore, (f ^ g)(30) = 2.

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To find the value of [tex](f ^ g)(30)[/tex], substitute 30 into both f(x) and g(x) to find f(30) and g(30). Then substitute g(30) into f(x) to find the final value.

In this mathematics question, you're asked to find the value of [tex](f ^ g)(30)[/tex]where [tex]f(x) = 2x - 8[/tex]and g(x) = square root of x - 5. To calculate (f ^ g)(x), we first need to find the value x in both functions f and g. So, let's find f(30) and g(30) first.

Firstly, substitute x = 30 into f(x). So, [tex]f(30) = 2*30 - 8 = 52[/tex]. Now, substitute x = 30 into g(x). So, g(30) = square root of (30 - 5) = square root of 25 = 5.

Now, f ^ g(30) means we substitute the output of g(30) = 5 into f(x) which gives us [tex]f(5) = 2*5 - 8 = 2.[/tex]

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Answer:

x = ± 3 and ± 2

Step-by-step explanation:

The given quadratic equation is

[tex](x^{2}-1)^{2}-11(x^{2}-1)+24=0[/tex]

To make this question easier we will consume ( x² -1 ) = a

a² - 11a + 24 = 0

Now we can factorize this equation easily

a²- 8a - 3a + 24 = 0

a (a-8) - 3 ( a-8) = 0

( a-3 ) ( a-8 ) = 0

Therefore, a - 3 = 0 ⇒ a = 3

or               a - 8 = 0 ⇒ a = 8

Now we put the value a

when a = 3    (x² - 1) = 3

                     x² = 3 + 1 = 4

                    x = √4

                       = ± 2

when a = 8   (x² - 1) = 8

                     x² = 9

                     x = √9

                        = ± 3

Therefore, x = ± 3 and ± 2 will be the answer.

To change 0.45 to a percent, we can move the decimal point 2 places to the right and add the percent sign.

If we move the decimal 2 places to the right, we will end up with 45 and then we will add the percent sign and it will become 45%

Therefore, 0.45 to a percent is 45%

Since the mean waiting time is 12 minutes, the median waiting time would also be 12 minutes.

Here's how we can find the median waiting time:

Since the mean waiting time is 12 minutes, it means that if you were to add up all the waiting times of different arrivals and divide by the number of arrivals, you would get 12 minutes.

Now, for the median waiting time, we're looking for the value that falls exactly in the middle of all the waiting times.

If the distribution is symmetrical, which is often the case with waiting times, the mean and median will be the same.

MAD tells us how far, on average, all values are from the middle. So, in the example weight of elk antler pairs at Mount St Helens NVM are, on average, 2 away from the middle. On the other hand, weight of elk antler pairs at Rocky Mountain NP are, on average, 1.5 away from the middle. So, we can assure that elk antler pairs at Rocky Mountain NP weighs more than elk antler pairs at Mount St Helens NVM.

Hence, the sample standard deviation is:

                         1.1832

The data is given by:

8, 8, 8, 5, 9, 8, 7, 6, 6, 7, 7, 7, 7, 9, 9.

The mean of these data points is given by:

[tex]Mean(x')=\dfrac{8+8+8+5+9+8+7+6+6+7+7+7+7+9+9}{15}\\\\i.e.\\\\Mean(x')=\dfrac{111}{15}\\\\i.e.\\\\Mean(x')=7.4[/tex]

Now,

x                x-x'                   (x-x')²

8            8-7.4=0.6              0.36

8            8-7.4=0.6              0.36

8            8-7.4=0.6              0.36    

5            5-7.4= -2.4             5.76

9            9-7.4=1.6                2.56

8            8-7.4=0.6              0.36

7            7-7.4= -0.4             0.16

6            6-7.4= -1.4              1.96

6            6-7.4= -1.4              1.96

7            7-7.4= -0.4             0.16

7            7-7.4= -0.4             0.16

7            7-7.4= -0.4             0.16

7            7-7.4= -0.4             0.16

9            9-7.4=1.6                2.56

9            9-7.4=1.6                2.56

                             ∑(x-x')²=19.69          

Now the variance of the sample population is given by:

[tex]Variance=\dfrac{\sum (x-x')^2}{n-1}[/tex]

where  n is the number of data points.

Here n= 15

Hence, n-1=14

Hence, we get:

[tex]Variance=\dfrac{19.6}{14}\\\\i.e.\\\\Variance=1.4[/tex]

We know that the standard deviation is the square root of the variance.

i.e.

[tex]Standard\ deviation=\sqrt{1.4}\\\\i.e.\\\\Standard\ deviation=1.1832[/tex]