Joe is a college football kicker. At a point about halfway through the season he had made only 7 out of 26 field goal kicks for his team. This gives him a really lousy success rate. His coach wants his success rate to rise to 49% by Joe kicking a series of consecutive field goals successfully. How many consecutive field goals would Joe have to kick, and make, for his success rate to rise to the level his coach wants?

The position of the particle at time t = 1 is x = 4 units.

To find the position of the particle at time t = 1, we need to integrate the given velocity function v(t) with respect to time from 0 to 1:

x(t) = ∫v(t)dt (from t = 0 to t = 1)

= ∫([tex]3t^2[/tex]+ 6t)dt (from t = 0 to t = 1)

= ([tex]t^3 + 3t^2[/tex]) (from t = 0 to t = 1)

[tex]= (1^3 + 3(1^2)) - (0^3 + 3(0^2))[/tex]

= 4

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The minimum sample size which is need to to create the given confidence interval is equal to 467.

Sample size n

z = z-score for the desired confidence level

From attached table,

For 85% confidence level, which corresponds to a z-score of 1.44.

Maximum error or margin of error E = 0.06

Population standard deviation σ = 0.9

Minimum sample size required to construct a 85% confidence interval with an error of no more than 0.06,

Use the formula,

n = (z / E)^2 × σ^2

Plugging in the values, we get,

⇒ n = (1.44 / 0.06)^2 × 0.9^2

⇒ n = 466.56

Rounding up to the next integer, we get a minimum sample size of 467.

Therefore, the minimum sample size required to construct a 85% confidence interval with an error of no more than 0.06 is 467.

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Answer: 4.5

Step-by-step explanation:

Using the Maclaurin polynomial of degree 6, the approximation for the integral ∫³√(1+x²)dx from 0 to 0.4 is ≈ 0.41721.

To find the Maclaurin polynomial of degree 6 for the function g(x) = ³√(1+x²), we will use the binomial series expansion:

(1+x)^(n) = 1 + nx + (n(n-1)x²)/2! + (n(n-1)(n-2)x³)/3! + ...

In our case, n = 1/3, and x = x²:

g(x) = (1+x²)^(1/3) = 1 + (1/3)x² - (1/9)(2/3)x⁴/2! + (1/27)(2/3)(-1/3)x⁶/3! + ...

Now, we can write the Maclaurin polynomial of degree 6:

g(x) ≈ 1 + (1/3)x² - (1/27)x⁴ + (2/729)x⁶

To approximate the integral, we can integrate the polynomial from 0 to 0.4:

∫(1 + (1/3)x² - (1/27)x⁴ + (2/729)x⁶)dx from 0 to 0.4 ≈ [x + (1/9)x³ - (1/135)x⁵ + (1/2187)x⁷] evaluated from 0 to 0.4

Now, plug in the limits:

≈ [0.4 + (1/9)(0.4³) - (1/135)(0.4⁵) + (1/2187)(0.4⁷)] - [0 + (1/9)(0³) - (1/135)(0⁵) + (1/2187)(0⁷)]

≈ 0.4 + 0.01778 - 0.00059 + 0.00002

≈ 0.41721

Thus, using the Maclaurin polynomial of degree 6, the approximation for the integral ∫³√(1+x²)dx from 0 to 0.4 is ≈ 0.41721.

This can be evaluated using basic integration techniques to get an approximate value of the integral. This method is useful for approximating integrals that cannot be solved exactly, and the accuracy of the approximation can be improved by using higher degree Maclaurin polynomials.

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Answer: 8 cups

Step-by-step explanation: 16 cups in a gallon, Half of 16 is 8. (16 divided by 2 = 8)

The integer that represents Andrews depth in ¼ of an hour is 60 feet.

Word problems are sentences describing a 'real-life' situation where a problem needs to be solved by way of a mathematical calculation e.g. calculation of length and depth.

If Andrew descends at a rate of 4 feet per minute and we want to find his depth in ¼ of an hour.

1/4 of an hour = (1/4 * 60) minutes = 15 minutes

Thus, the integer that represents Andrews depth in ¼ of an hour will be:

(4 feet per minute) * (15 minutes) = 60 feet

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The distance between the airplane and the tower is approximately 1864.5 meters.

To solve this problem, we can use trigonometry. Let's draw a diagram to help us visualize the situation:

```
   T
  /|
 / |
/  | 500m
/a  |
--------
  x
```

In this diagram, "T" represents the control tower, "a" represents the airplane, and "x" represents the distance between them. We know that the height of the airplane is 500m, and the angle of depression from the tower to the airplane is 15 degrees. This means that the angle between the horizontal ground and the line from the tower to the airplane is also 15 degrees.

Using trigonometry, we can set up the following equation:

```
tan 15 = 500 / x
```

We can solve for "x" by multiplying both sides by "x" and then dividing by tan 15:

```
x = 500 / tan 15
```

Using a calculator, we can find that tan 15 is approximately 0.2679. Therefore:

```
x = 500 / 0.2679
x ≈ 1864.5m
```

So the distance between the airplane and the tower is approximately 1864.5 meters.

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Answer:

Constant

Step-by-step explanation:

What is an exponential function?

An exponential function is a function with the general form y = abx, a ≠ 0, b is a positive real number and b ≠ 1. In an exponential function, the base b is a constant. The exponent x is the independent variable where the domain is the set of real numbers.

In this case, y=ab^x

where 0.983 is in our b term, which gives the meaning that number is our constant in this exponential function.

The bases could be circled 40 times in 50 minutes at the same rate.

For this issue's solution, let's use unit rates.

In order to calculate the unit rate,

Considering that the player went 12 times around the bases in 15 minutes, the unit rate is 12/15, =  0.8 times per minute.

In a minute, the player would have circled the bases 0.8 times. By dividing the unit rate by the number of minutes, we can calculate how many times the bases could be circled in 50 minutes:

50 minutes x  0.8 times each minute = 40 times.

Therefore, the bases could be circled 40 times in 50 minutes at the same rate.

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To prove that the expression (36^5−6^9)(38^9−38^8) is divisible by 30, we need to show that it is divisible by both 2 and 3.

First, we can factor out a 6^9 from the first term:

(36^5−6^9)(38^9−38^8) = 6^9(6^10-36^5)(38^9-38^8)

Notice that 6^10 can be written as (2*3)^10, which is clearly divisible by both 2 and 3. Also, 36 is divisible by 3, so 36^5 is divisible by 3^5. Thus, we can write:

6^9(6^10-36^5) = 6^9(2^10*3^10 - 3^5*2^10) = 6^9*2^10*(3^10 - 3^5)

Since 2^10 is divisible by 2, and 3^10 - 3^5 is clearly divisible by 3, the whole expression is divisible by both 2 and 3, and therefore divisible by 30.

To prove that the expression is divisible by 37, we can use Fermat's Little Theorem. Fermat's Little Theorem states that if p is a prime number and a is any positive integer not divisible by p, then a^(p-1) is congruent to 1 modulo p, which can be written as a^(p-1) ≡ 1 (mod p).

In this case, p = 37, and 36 is not divisible by 37. Therefore, by Fermat's Little Theorem:

36^(37-1) ≡ 1 (mod 37)

Simplifying the exponent gives:

36^36 ≡ 1 (mod 37)

Similarly, 38 is not divisible by 37, so:

38^(37-1) ≡ 1 (mod 37)

Simplifying the exponent gives:

38^36 ≡ 1 (mod 37)

Now we can use these congruences to simplify our expression:

(36^5−6^9)(38^9−38^8) ≡ (-6^9)(-1) ≡ 6^9 (mod 37)

We know that 6^9 is divisible by 3, so we can write:

6^9 = 2^9*3^9

Since 2 and 37 are relatively prime, we can use Euler's Totient Theorem to simplify 2^9 (mod 37):

2^φ(37) ≡ 2^36 ≡ 1 (mod 37)

Therefore:

2^9 ≡ 2^9*1 ≡ 2^9*2^36 ≡ 2^(9+36) ≡ 2^45 (mod 37)

Now we can simplify our expression further:

6^9 ≡ 2^45*3^9 ≡ (2^5)^9*3^9 ≡ 32^9*3^9 (mod 37)

Notice that 32 is congruent to -5 modulo 37, since 32+5 = 37. Therefore:

32^9 ≡ (-5)^9 ≡ -5^9 ≡ -1953125 ≡ 2 (mod 37)

So:

6^9 ≡ 2*3^9 ≡ 2*19683 ≡ 39366 ≡ 0 (mod 37)

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To find three positive integers x, y, and z that satisfy the given conditions, we need to use the concept of maximizing a function subject to certain conditions. Solving for y and z, we have y = 15 and z = 16.

In this case, we want to maximize the function P= xy^2z, subject to the condition that the sum of x, y, and z is 32.
To maximize P, we need to find the values of x, y, and z that make P as large as possible. One way to do this is to use the method of Lagrange multipliers, which involves finding the critical points of a function subject to a constraint.
In this case, we have the function P= xy^2z and the constraint x+y+z=32. Using Lagrange multipliers, we can set up the following equations:
∂P/∂x = λ∂(x+ y+ z)/∂x
y^2z = λ
∂P/∂y = λ∂(x+ y+ z)/∂y
2xyz = λ
∂P/∂z = λ∂(x+ y+ z)/∂z
xy^2 = λ
x+y+z=32
Solving these equations simultaneously, we get:
y^2z/x = 2xyz/y = xy^2/z = λ
Simplifying, we get:
y^2z/x = 2yz = xy^2/z
Rearranging, we get:
x = 2y^3/z
y = (x/2z)^(1/3)
z = (x/4y^2)^(1/3)
Substituting these expressions for x, y, and z into the constraint x+y+z=32, we get:
2y^3/z + (x/2z)^(1/3) + (x/4y^2)^(1/3) = 32
Solving this equation for x, y, and z, we get:
x = 16
y = 4
z = 2
Therefore, the three positive integers x, y, and z that satisfy the given conditions are x=16, y=4, and z=2. These values make P= xy^2z a maximum, since any other values of x, y, and z that satisfy the constraint x+y+z=32 would yield a smaller value of P.

To find three positive integers x, y, and z that satisfy the given conditions, we need to consider the following:
1. The sum of x, y, and z is 32: x + y + z = 32
2. The product P = xy^2z is a maximum.
First, let's express z in terms of x and y using the sum condition:
z = 32 - x - y
Now, substitute this expression for z into the product P:
P = xy^2(32 - x - y)
To maximize P, we should make y as large as possible, since it has the largest exponent in the product formula. Let's allocate the majority of the remaining sum to y. For example, if x = 1, we get:
1 + y + z = 32
Solving for y and z, we have y = 15 and z = 16. Now let's check the product:
P = (1)(15^2)(16) = 3600
This is one possible solution for x, y, and z that gives a maximum product P with the given conditions. The three positive integers are x = 1, y = 15, and z = 16, and the maximum product P = 3600.

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The best estimate of the number of ants in Mrs. Brown's yard is 15,116,596.

What is arithmetic sequence?

An arithmetic sequence is a sequence of numbers in which each term after the first is found by adding a fixed constant number, called the common difference, to the preceding term.

Mrs. Brown's yard has three ant hills, each with a different number of ants. To estimate the total number of ants in the yard, we simply add up the number of ants in each hill.

The first hill has 4,867,190 ants, the second has 6,256,304, and the third has 3,993,102. When we add these numbers together, we get a total of 15,116,596 ants in Mrs. Brown's yard. Of course, this is just an estimate, as there may be other ant hills or individual ants scattered around the yard.

However, this calculation gives us a good approximation of the number of ants in the yard based on the information given.

To estimate the total number of ants in Mrs. Brown's yard, we can add up the number of ants in each of the three ant hills:4,867,190 + 6,256,304 + 3,993,102 = 15,116,596.

Therefore, the best estimate of the number of ants in Mrs. Brown's yard is 15,116,596.

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Answerx = 131.25

Step-by-step explanation:f x varies directly as T, then we can use the formula for direct variation:

x = kT

where k is the constant of proportionality.

To find k, we can use the given values:

x = 105 when T = 400

105 = k(400)

k = 105/400

k = 0.2625

Now that we have the value of k, we can use the formula to find x when T = 500:

x = kT

x = 0.2625(500)

x = 131.25

Therefore, when T = 500, x is equal to 131.25.

The total cost after discount and tax is $46.28.

To find the total cost after the discount and tax, we need to first find the subtotal before tax.

Fred bought two pairs of jeans for $17.99 each, so the cost of the jeans is $17.99 x 2 = $35.98.

He also bought three shirts at $5.99 each, so the cost of the shirts is $5.99 x 3 = $17.97.

The pack of socks costs $3.

The subtotal before discount is $35.98 + $17.97 + $3 = $57.95.

With the 25% off coupon, Fred gets a discount of $57.95 x 0.25 = $14.49.

The new subtotal after discount is $57.95 - $14.49 = $43.46.

Finally, we need to add the 6.5% sales tax.

The sales tax is $43.46 x 0.065 = $2.82.

The total cost after discount and tax is $43.46 + $2.82 = $46.28.

Therefore, the closest answer is C: $45.49.

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The trapezoidal approximation for | Vx dx using a regular partition with n=6 is approximately 0.1901.

To compute the trapezoidal approximation for | Vx dx using a regular partition with n=6, we can use the formula:

Tn = (b-a)/n * [f(a)/2 + f(x1) + f(x2) + ... + f(xn-1) + f(b)/2]

where Tn is the trapezoidal approximation, n=6 is the number of partitions, a and b are the limits of integration, and x1, x2, ..., xn-1 are the partition points.

In this case, we have | Vx dx as the function to integrate. Since there are no given limits of integration, we can assume them to be 0 and 1 for simplicity.

So, a=0 and b=1, and we need to find the values of f(x) at x=0, 1/6, 2/6, 3/6, 4/6, and 5/6 to use in the formula.

We can calculate these values as follows:

f(0) = | V0 dx = 0

f(1/6) = | V1/6 dx = V(1/6) - V(0) = sqrt(1/6) - 0 = 0.4082

f(2/6) = | V2/6 dx = V(2/6) - V(1/6) = sqrt(2/6) - sqrt(1/6) = 0.2317

f(3/6) = | V3/6 dx = V(3/6) - V(2/6) = sqrt(3/6) - sqrt(2/6) = 0.1547

f(4/6) = | V4/6 dx = V(4/6) - V(3/6) = sqrt(4/6) - sqrt(3/6) = 0.1104

f(5/6) = | V5/6 dx = V(5/6) - V(4/6) = sqrt(5/6) - sqrt(4/6) = 0.0849

Now we can substitute these values in the formula and simplify:

T6 = (1-0)/6 * [0/2 + 0.4082 + 0.2317 + 0.1547 + 0.1104 + 0.0849/2]
   = 0.1901

Therefore, the trapezoidal approximation for | Vx dx using a regular partition with n=6 is approximately 0.1901.

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The  expected value and standard error for the number of winning plays is $3.1 and $ 2.21.

The population mean's likelihood to differ from a sample mean is indicated by the standard error of a mean, and simply standard error.

It reveals how much what the sample mean will change if a study were to be repeated with fresh samples drawn from a single population.

Chance of winning $3 = 2/10

chance of losing $2 = 3/10

chance of losing $1 = 5/10

Average of tickets. = - $2.50

SD of tickets = $1.80.

The box model for net gain has 2 tickets labeled $3, 3 tickets labeled $2, 5 tickets labeled

a) Expected value for the net gain.

The Expected value for net = ∑ x.p(x)

Here

can take value $1, $2 and $3

Here p(x) us the probability of winning respectively.

So, Now, Expected gain is,

(2 x 3 x 2/10) + (3 x -2 x 7/10) + (5 x -1 x 5/10)

= 12/10 - 18/10 - 25/10

= -31/10 = -$3.1.

b) Standard error of the net gain,

S.D = [tex]\sqrt{E(x^2) - [E(x)]^2}[/tex]

Now E(x²) = (2 x 3² x 2/10) + (3 x -2² x 7/10) + (5 x -1² x 5/10)

= 36/10 + 84/10 + 25/10 = 145/10

= $ 14.5

SD = [tex]\sqrt{14.5 - 3.1}\\[/tex]

SD = $ 2.21

c) Chance that the net gain is $15

P(X=15) = (z = 15-(-2.50)/1.80

= P(z=9.72) = 0.99.

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The remaining distance after 33 minutes of driving = 58.8 miles.

Here, the slope of a linear function the remaining distance to drive (in miles) is -0.85

For this situation, we can write a linear equation as,

remaining distance = (slope)(drive time) + (intercept)

remaining distance = -0.85(drive time) + (intercept)      

y =  -0.85x + c        ..........(1)

where y represents the remaining distance

x is the drive time

and c is the y-intercept

Here, Laura has 52 miles remaining after 41 minutes of driving.

i.e., x = 41 and y = 52

Substitute these values in equation (1)

52 =  -0.85(41) + c    

c = 52 + 34.85

c = 86.85

So, equation (1) becomes,

y =  -0.85x + 86.85

Now, we need to find the remaining distance after 33 minutes of driving.

i.e., the value of y for x = 33

y =  -0.85(33) + 86.85

y =  -28.05 + 86.85        

y = 58.8

This is the remaining distance 58.8 miles.

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The best estimate for the volume of the tank in cubic feet is 5.5 cubic feet.

The volume of the tank is:

V = l x w x h

where l is the length, w is the width, and h is the height.

Substituting the given values, we get:

V = 25 x 34.5 x 11 = 9547.5 cubic inches

To convert cubic inches to cubic feet, we divide by (12 x 12 x 12), since there are 12 inches in a foot and 12 x 12 x 12 cubic inches in a cubic foot:

V = 9547.5 / (12 x 12 x 12) cubic feet

V ≈ 5.5 cubic feet

Therefore, 5.5 cubic feet is the best estimate for the tank's cubic foot capacity.

Hence , the volume of the rectangular glass tank is 5.490 feet³

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In general, it is best to determine the equation of the line of best fit exactly rather than relying on estimation by inspection. This is because an exact equation allows for more precise predictions and calculations.

Estimation by inspection can be useful in situations where the data is relatively simple and a rough estimate is sufficient. However, in more complex datasets, it is important to use statistical methods to determine the line of best fit accurately.

For example, in some cases, it may be more important to prioritize the accuracy of the slope of the line over the accuracy of the intercept. In such cases, certain methods, such as minimizing the sum of the squares of the vertical deviations, may be more appropriate than others.

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lets assume.

Algebraic expression: c - 2.5. The variable is c, which represents Julie's initial weight in kilograms.

Algebraic equation: Jhon's height = 2m + 5. The variables are m, which represents Peter's height in centimeters, and the height of Jhon, which is represented by the equation.

Algebraic equation: Ador's age = 3(Emy's age - d). The variables are Ador's age and Emy's age, which is d years less than 9.

Algebraic expression: n + 7. The variable is n, which represents Jupiter's current age in years.

Algebraic equation: Anna's age = 3p + 4. The variables are p, which represents Anna's sister's age in years, and Anna's age, which is represented by the equation.

SO ANNA  current age is P=3+4

and p=7

Anna's age = 3p + 4, where p is the age of Anna's sister in years, and Anna is 4 years older than thrice the age of her sister.

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The expression for the calculation of adding 8 to the sum of 23 and 10 is 8 + (23 + 10)

To calculate expression parentheses the sum of 23 and 10, we add them together, which gives us 33. Then, we add 8 to that result, giving us a final answer of 41. So, the expression 8 + (23 + 10) equals 41.

This expression follows the order of operations, which states that we should first perform the addition inside the parentheses and then add the result to 8.

expressions are made up of numbers and symbols, and they represent a mathematical relationship or operation. In this case, the expression includes addition and parentheses, which tell us to perform the addition inside them first. The parentheses clarify which numbers should be added together first before adding 8.

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The number of type 2 cabinets produced last week is 80. The number of type 2 cabinets produced last week exceeded the number of type 1 cabinets produced during the week by 50.

Using the system of equations given, we can solve for the number of type 1 and type 2 cabinets produced.

x + y = 110 represents the total number of cabinets produced, where x is the number of type 1 cabinets and y is the number of type 2 cabinets produced.

y = 2x + 20 represents the relationship between the number of type 1 and type 2 cabinets produced. This equation tells us that the number of type 2 cabinets produced exceeds twice the number of type 1 cabinets produced by 20.

To solve for y, we substitute the value of y from the second equation into the first equation:

x + (2x + 20) = 110

Simplifying this equation:

3x + 20 = 110

3x = 90

x = 30

Therefore, the number of type 1 cabinets produced last week is 30.

To find the number of type 2 cabinets produced, we substitute x = 30 into the second equation:

y = 2x + 20 = 2(30) + 20 = 80

The number of type 2 cabinets produced last week exceeds the number of type 1 cabinets produced during the week by:

80 - 30 = 50.

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To solve this problem, we first need to understand what "at most 4 of them were successes" means. This includes the cases where there are 0, 1, 2, 3, or 4 successful serves out of the 6 selected.

We can use the table to find the probabilities for each of these cases.

For 0 successful serves, the probability is 0.0002.

For 1 successful serve, the probability is 0.004.

For 2 successful serves, the probability is 0.033.

For 3 successful serves, the probability is 0.132.

For 4 successful serves, the probability is 0.297.

To find the probability of at most 4 successful serves, we add up these probabilities:
[tex]0.0002 + 0.004 + 0.033 + 0.132 + 0.297 = 0.466[/tex]

So the probability of at most 4 successful serves is 0.466.

Therefore, the answer is 0.466 and it is found by adding up the probabilities for the cases where there are 0, 1, 2, 3, or 4 successful serves out of the 6 selected from the table.

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Minimizing the sum of squared residuals.

The OLS (Ordinary Least Squares) estimator is derived by minimizing the sum of squared residuals. This method aims to find the line of best fit that minimizes the vertical distance between the observed data points (Yi) and the predicted values on the regression line. The residuals represent the differences between the observed values and the predicted values.

By minimizing the sum of squared residuals, the OLS estimator ensures that the line fits the data as closely as possible. This approach is based on the principle of least squares, which seeks to find the parameters that minimize the overall discrepancy between the observed data and the predicted values.

Connecting the Yi corresponding to the lowest Xi observation with the Yi corresponding to the highest Xi or ensuring that the standard error of the regression equals the standard error of the slope estimator are not the steps involved in deriving the OLS estimator. The OLS method specifically focuses on minimizing the sum of squared residuals.

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To use Lagrange multipliers, we need to define the Lagrangian function:

L(x, y, z, λ) = xyz + λ(x + y + z - 9)

Now, we need to find the partial derivatives of L with respect to x, y, z, and λ and set them equal to 0:

∂L/∂x = yz + λ = 0
∂L/∂y = xz + λ = 0
∂L/∂z = xy + λ = 0
∂L/∂λ = x + y + z - 9 = 0

From the first three equations, we can see that:

yz = -λ
xz = -λ
xy = -λ

Multiplying these equations together, we get:

(xyz)^2 = (-λ)^3

Substituting λ = -yz into the fourth equation, we get:

x + y + z - 9 = 0

Substituting λ = -yz into the first equation and solving for x, we get:

x = -λ/yz = (yz)^2/(-yz) = -y^2z^2

Similarly, we can solve for y and z:

y = -x^2z^2
z = -x^2y^2

Substituting these expressions into the constraint equation, we get:

(-y^2z^2) + (-x^2z^2) + (-x^2y^2) - 9 = 0

Simplifying and solving for xyz, we get:

xyz = sqrt(9/(x^2 + y^2 + z^2))

To maximize xyz, we need to minimize x^2 + y^2 + z^2. Therefore, we can set:

x^2 + y^2 + z^2 = 3

Substituting this into the expressions for x, y, and z, we get:

x = -y^2z^2
y = -x^2z^2
z = -x^2y^2

Substituting these expressions into xyz, we get:

xyz = sqrt(9/3) = 3

Therefore, the maximum value of f(x, y, z) = xyz subject to the constraint x + y + z - 9 = 0 is 3.
To solve this problem using Lagrange multipliers, we first set up the Lagrangian function L(x, y, z, λ) with the constraint function g(x, y, z) = x + y + z - 9.

L(x, y, z, λ) = f(x, y, z) - λ(g(x, y, z))

L(x, y, z, λ) = xyz - λ(x + y + z - 9)

Now we take the partial derivatives with respect to x, y, z, and λ, and set them equal to 0:

∂L/∂x = yz - λ = 0
∂L/∂y = xz - λ = 0
∂L/∂z = xy - λ = 0
∂L/∂λ = x + y + z - 9 = 0 (the constraint)

From the first three equations, we get:

yz = xz = xy

Since x, y, and z are positive, we can divide the first two equations:

y/z = x/z => y = x
x/z = y/z => x = y

So x = y = z. Now we can use the constraint equation:

x + x + x - 9 = 0 => 3x = 9 => x = 3

Thus, x = y = z = 3. Now we can find the maximum value of f(x, y, z):

f(3, 3, 3) = 3 * 3 * 3 = 27

So the maximum value of f(x, y, z) = xyz subject to the constraint x + y + z - 9 = 0 is 27, and this occurs at the point (3, 3, 3).

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Answer:

Step-by-step explanation:

a. To find the time it takes for the stone to hit the floor, we can use the formula t = sqrt(2h/g), where h is the height of the table and g is the acceleration due to gravity. Plugging in the values, we get:

t = sqrt(2(1.5 m)/9.8 m/s^2) = 0.55 seconds.

b. To find the horizontal distance traveled by the stone, we can use the formula d = vt, where v is the initial velocity of the stone and t is the time it takes to hit the floor. Plugging in the values, we get:

d = (0.80 m/s) * (0.55 s) = 0.44 meters.

Therefore, the stone will hit the floor after 0.55 seconds and will travel 0.44 meters from the table.

The required area of the shaded part of the circle is 50.24 sq. cm

A circle of radius r has an area of r2 in geometry. Here, the Greek letter  denotes the constant ratio of a circle's diameter to circumference, which is roughly equivalent to 3.14159.

Given data:

Radius of small circle = 6/2 = 3 cm

Radius of big circle = 10/2 = 5 cm

then.

Area of shaded part = area of big circle - area of small circle

Area of shaded part = π(5)² -  π(3)²

Area of shaded part = 25π - 9π

Area of shaded part = 16π

Area of shaded part = 50.24 sq. cm

Thus, required area of the shaded part of the circle is 50.24 sq. cm

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The volume of the cone is 57 cubic inches. To find the volume of the cone, we use the formula: V = (1/3)π[tex]r^{2}[/tex]h, where r is the radius of the cone, h is the height of the cone, and π is approximately 3.14.

Given that the radius of the cone is 3 inches and the height is 6 inches, we can substitute these values into the formula and solve for V: V = (1/3)π([tex]3^{2}[/tex])(6), V = (1/3)π(9)(6), V = (1/3)(3.14)(54), V = 56.52 cubic inches

Rounding to the nearest whole number, the volume of the cone is 57 cubic inches.

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There exists at least one c in the open interval (a, b) such that f'(c) = 0.

There are 3 distinct points where p'(x) = 0, which corresponds to the 3 distinct real solutions of the given equation.

To prove that the given equation has exactly 3 distinct real solutions, let's follow the steps mentioned in the question.

First, consider the polynomial p(x) = (x-a)(x-b)(x-c)(x-d). Since a, b, c, and d are distinct real numbers, p(x) has 4 distinct real roots, namely a, b, c, and d.

Now, let's find the derivative p'(x) using logarithmic differentiation. Taking the natural logarithm of both sides, we have:

[tex]ln(p(x)) = ln((x-a)(x-b)(x-c)(x-d))[/tex]

Differentiating both sides with respect to x, we get:

[tex]p'(x)/p(x) = 1/(x-a) + 1/(x-b) + 1/(x-c) + 1/(x-d)[/tex]

Multiplying both sides by p(x) and simplifying, we have:

[tex]p'(x) = (x-b)(x-c)(x-d) + (x-a)(x-c)(x-d) + (x-a)(x-b)(x-d) + (x-a)(x-b)(x-c)[/tex]

Now, we apply Rolle's Theorem, which states that if a function is continuous on the closed interval [a, b], differentiable on the open interval (a, b), and f(a) = f(b), then there exists at least one c in the open interval (a, b) such that f'(c) = 0.

Since p(x) has 4 distinct real roots, there must be 3 intervals between these roots where the function p(x) satisfies the conditions of Rolle's Theorem. Therefore, there are 3 distinct points where p'(x) = 0, which corresponds to the 3 distinct real solutions of the given equation.

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