what are the degrees of freedom for the f test on whether hours affects salary? a. (1, 49) b. (50, 1) c. (1, 50) d. (49, 1)

Answer:

x = 48

Step-by-step explanation:

the sum of the 3 angles in a triangle = 180°

sum the 3 angles and equate to 180

x + 105 + 27 = 180

x + 132 = 180 ( subtract 132 from both sides )

x = 48

If the severity of Type I error is high, meaning that it would be very costly or harmful to falsely reject the null hypothesis, then a more stringent alpha level would be appropriate. In this case, option b, 0.001, would be the most appropriate significance level as it would minimize the chance of a Type I error occurring.

An appropriate significance level (alpha level) for a hypothesis test where the severity of Type I error is high would be a lower alpha value. This is because a lower alpha level reduces the likelihood of committing a Type I error (incorrectly rejecting the null hypothesis).

In this case, the appropriate significance level among the given options is:

b) 0.001

A lower alpha level like 0.001 indicates that there is a smaller chance of committing a Type I error, making it more suitable when the severity of Type I error is high.

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As f'(10) > 0. Thus, this means that f is increasing function on the interval (9, ∞).

To determine the intervals on which f is increasing, we need to look at the sign of the derivative f'(x). Recall that if f'(x) > 0, then f is increasing on the interval, and if f'(x) < 0, then f is decreasing on the interval.

First, we need to find the critical points of f. These are the values of x where f'(x) = 0 or does not exist. In this case, we see that f'(x) = 0 when x = 4, -8, and 9. So the critical points are x = 4, -8, and 9.

Next, we need to test the intervals between these critical points to see where f is increasing. We can do this by choosing test points within each interval and plugging them into f'(x).

For x < -8, we can choose a test point of -10. Plugging this into f'(x), we get:
f'(-10) = (-14)^8 * (-2)^5 * (-19)^6

All of these factors are negative, so f'(-10) < 0. This means that f is decreasing on the interval (-∞, -8).
For -8 < x < 4, we can choose a test point of 0. Plugging this into f'(x), we get:
f'(0) = (-4)^8 * (8)^5 * (-9)^6

The first and third factors are positive, while the second factor is negative. Thus, f'(0) < 0, so f is decreasing on the interval (-8, 4).
For 4 < x < 9, we can choose a test point of 6. Plugging this into f'(x), we get:
f'(6) = (2)^8 * (14)^5 * (-3)^6

All of these factors are positive, so f'(6) > 0. This means that f is increasing on the interval (4, 9).

Finally, for x > 9, we can choose a test point of 10. Plugging this into f'(x), we get:
f'(10) = (6)^8 * (18)^5 * (1)^6

All of these factors are positive, so f'(10) > 0. This means that f is increasing on the interval (9, ∞).
Putting all of this together, we see that f is increasing on the intervals (4, 9) and (9, ∞).

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The bottom right graph shows the line of best fit for the data.

For a data-set, the definition of a residual is that it is the difference of the actual output value by the predicted output value, that is:

Residual = Observed - Predicted.

Hence the graph of the line of best fit should have the smallest possible residual values, meaning that the points on the scatter plot are the closest possible to the line.

For this problem, we have that the bottom right graph has the smaller residuals, hence it shows the line of best fit for the data.

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The half-life of the goo is approximately 18.75 minutes. The formula for g(t), the amount of goo remaining at time t, is g(t) = 132 * (1/2)^(t/18.75).

To find the half-life of the goo, we can use the formula for exponential decay: A(t) = A0 * (1/2)^(t/h), where A(t) is the amount of radioactive substance at time t, A0 is the initial amount, h is the half-life, and t is time. We are given A0 = 132 grams, A(75) = 2.0625 grams, and we need to solve for h. Plugging in these values, we get:

2.0625 = 132 * (1/2)^(75/h)

Solving for h, we get h ≈ 18.75 minutes.

The formula for g(t) is g(t) = A0 * (1/2)^(t/h). Plugging in A0 = 132 and h = 18.75, we get g(t) = 132 * (1/2)^(t/18.75). This formula gives us the amount of goo remaining at time t, where t is measured in minutes.

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The norm of u is 276.46 and the distance between u and v is 259.98 in M22 with the standard inner product.

To find the norm ||u|| of the vector u=[39 276] in M22 with the standard inner product, we use the formula:

||u|| = sqrt(<u,u>)

where <u,u> is the dot product of u with itself.

<u,u> = (39 * 39) + (276 * 276) = 76461

Therefore, ||u|| = sqrt(76461) = 276.46 (rounded to two decimal places).

To find the distance d(u,v) between vectors u=[39 276] and v=[-64 19] in M22 with the standard inner product, we use the formula:

d(u,v) = sqrt(<u-v,u-v>)

where <u-v,u-v> is the dot product of the difference between u and v with itself.

<u-v,u-v> = (39 - (-64))^2 + (276 - 19)^2 = 12769 + 54756 = 67525

Therefore, d(u,v) = sqrt(67525) = 259.98 (rounded to two decimal places).

Therefore, the norm of u is 276.46 and the distance between u and v is 259.98 in M22 with the standard inner product.

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Therefore, According to the given information f(5) = 35.

To find the value of f(5), we can use the information given about f(2) and the integral of f′(t) from 2 to 5. Here's a step-by-step explanation:
1. We know that f(2) = 14.
2. We also know that the integral of f′(t) from 2 to 5 is equal to 21. This represents the accumulated change in the function f(t) from 2 to 5.
3. Since f′(t) is continuous, we can use the Fundamental Theorem of Calculus to relate the integral of f′(t) to the function f(t).
4. The Fundamental Theorem of Calculus states that the integral of f′(t) from 2 to 5 is equal to f(5) - f(2).
5. Plugging in the known values, we have 21 = f(5) - 14.
6. Solve for f(5): f(5) = 21 + 14.

Therefore, According to the given information f(5) = 35.

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The equations of the functions are f(x) = 200(2.25)ˣ and f(x) = 100(0.84)ˣ

The values of a and b are 8 and 4.2

For problem card 1

An exponential function is represented as

f(x) = abˣ

Where

a = y-intercept

b = rate

Using the data card, we have

a = 200

So, we have

y = 200bˣ

Solving for b, we have

200b² = 1012.5

b² = 5.0625

b = 2.25

So, the function is f(x) = 200(2.25)ˣ

For problem card 2

An exponential function is represented as

f(x) = abˣ

Where

a = y-intercept

b = rate

Using the data card, we have

a = 100

So, we have

y = 100bˣ

Solving for b, we have

[tex]100b^{\frac14} = 50[/tex]

[tex]b^{\frac14} = 0.5[/tex]

b = 0.84

So, the function is f(x) = 100(0.84)ˣ

An exponential function is represented as

f(x) = abˣ

Where

a = y-intercept

b = rate

So, we have

a = 8

So, we have

y = 8bˣ

Solving for b, we have

b² = 18

b = 4.2

Hence, the values of a and b are 8 and 4.2

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So, the volume of the largest rectangular box that can be inscribed in the given ellipsoid is 384 cubic units.

To find the volume of the largest rectangular box inscribed in the ellipsoid (x^2/4) + (y^2/16) + (z^2/36) = 1, we can consider the semi-axes of the ellipsoid as the lengths of the rectangular box.

The equation of the ellipsoid can be rewritten as:

x^2/2^2 + y^2/4^2 + z^2/6^2 = 1

The semi-axes of the ellipsoid are given by (a, b, c), where a = 2, b = 4, and c = 6.

For a rectangular box inscribed in the ellipsoid, the length, width, and height of the box would be twice the semi-axes of the ellipsoid, i.e., (2a, 2b, 2c).

Therefore, the dimensions of the largest rectangular box are (4, 8, 12).

The volume of a rectangular box is given by the product of its dimensions. Hence, the volume of the largest rectangular box inscribed in the ellipsoid is:

Volume = (4)(8)(12)

= 384 cubic units

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Answer:

Angle PRQ = 28 degrees.

Step-by-step explanation:

Angles P and Q are the same! That's bc this is an isosceles triangle.

The total of all 3 angles = 180. So to find R, subtract the other 2 angles from 180.

So 180-76-76 = 28. That's angle R

The volume of the solid is (11π/3) cubic units.

We have,

To find the volume of the solid obtained by rotating the region bounded by y = x and y = x^2 about the line x = -3, we can use the method of cylindrical shells.

The formula for the volume using cylindrical shells is given by:

V = 2π ∫ [a, b] x h(x) dx,

where [a, b] is the interval of integration, x represents the variable of integration, and h(x) represents the height of the shell at each value of x.

In this case, we want to rotate the region bounded by y = x and y = x² about the line x = -3.

Since we are rotating about a vertical line, the height of the shell at each value of x will be given by the difference between the x-coordinate of the curve and the line of rotation:

h(x) = (x - (-3)) = x + 3.

To find the interval of integration, we need to determine the x-values where the two curves intersect.

Setting x = x², we have:

x = x²,

x² - x = 0,

x (x - 1) = 0.

This gives us two intersection points: x = 0 and x = 1.

Therefore, the interval of integration is [0, 1].

Now we can set up the integral to find the volume:

V = 2π ∫ [0, 1] x (x + 3) dx.

Evaluating this integral, we have:

V = 2π ∫ [0, 1] (x² + 3x) dx

= 2π [x³/3 + (3/2)x²] evaluated from 0 to 1

= 2π [(1/3 + 3/2) - (0/3 + 0/2)]

= 2π [(2/6 + 9/6) - 0]

= 2π (11/6)

= (22π/6)

= (11π/3).

Therefore,

The volume of the solid is (11π/3) cubic units.

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Thus, the general solution is y(x) = -x + 2x^3 + c₁x + c₂x^3.

To find the general solution of the differential equation x^2y'' + 3xy' + 3xy = 4x^7, we can use the method of variation of parameters.

Given that {x, x^3} is a fundamental set of solutions of the homogeneous equation x^2y'' - 3xy' + 3y = 0, we can use these solutions to find the particular solution.

Let's assume the particular solution has the form y_p = u(x)x + v(x)x^3, where u(x) and v(x) are unknown functions.

Differentiating y_p:

y_p' = u'x + u + v'x^3 + 3v(x)x^2

Differentiating again:

y_p'' = u''x + 2u' + v''x^3 + 6v'x^2 + 6v(x)x

Substituting these derivatives into the original differential equation, we have:

x^2(u''x + 2u' + v''x^3 + 6v'x^2 + 6v(x)x) + 3x(u'x + u + v'x^3 + 3v(x)x^2) + 3x(u(x)x + v(x)x^3) = 4x^7

Simplifying and grouping like terms:

x^3(u'' + 3v') + x^2(2u' + 3v'' + 3v) + x(u + 3v' + 3v) + (2u + v) = 4x^5

Setting the coefficients of each power of x to zero, we get the following system of equations:

x^3: u'' + 3v' = 0

x^2: 2u' + 3v'' + 3v = 0

x^1: u + 3v' + 3v = 0

x^0: 2u + v = 4

Solving this system of equations, we find:

u = -1

v = 2

Therefore, the particular solution is y_p = -x + 2x^3.

The general solution of the differential equation x^2y'' + 3xy' + 3xy = 4x^7 is given by the sum of the particular solution and the homogeneous solutions:

y(x) = y_p + c₁x + c₂x^3

where c₁ and c₂ are arbitrary constants.

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Therefore, p(1) = -1, p(2) = -1, p(3) = -2, and p(4) = 0. The recurrence relation is given by p(n) = 0 if n = 0 and [p(n-1)]^2 - n if n > 0.

We can use this to compute p(1), p(2), p(3), and p(4) as follows:

p(1) = [p(0)]^2 - 1 = 0^2 - 1 = -1

p(2) = [p(1)]^2 - 2 = (-1)^2 - 2 = -1

p(3) = [p(2)]^2 - 3 = (-1)^2 - 3 = -2

p(4) = [p(3)]^2 - 4 = (-2)^2 - 4 = 0

Therefore, p(1) = -1, p(2) = -1, p(3) = -2, and p(4) = 0.

To compute p(n) for larger values of n, we would need to use the recurrence relation repeatedly, plugging in the value of p(n-1) each time. However, it is worth noting that the recurrence relation leads to a sequence that grows very quickly in magnitude,

as each term is the square of the previous term minus a constant. Therefore, the values of p(n) for large values of n will be very large (in absolute value), and it may be difficult to compute them explicitly.

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The set of all real numbers for which the function is defined is the domain of the function.

We have the function is:

h(x) = −log(3x−8) + 5

The logarithmic function is defined only for real numbers that are greater than 0. Hence, this implies that (3x-8) must be greater than 0.

=> 3x - 8 > 0

=> 3x > 8

=> x > 8/3

Thus, the domain of the given function is all real numbers that are greater than 8/3.

Domain will be in interval is:

(8/3, infinity)

The values of x for which the function, f(x) is undefined and the limit of the function does not exist is the vertical asymptote of a function.

The given function is undefined when 3x-2 will be equal to 0.

The equation will be in the form and solve for 'x'.

3x - 8  = 0

3x = 8

x = 8/3

The value of x is 8/3.

Therefore, the vertical asymptote of the given function is x=8/3.

Find the limiting value of the given function when x approaches the vertical asymptote,

h(x) = -log(3x - 8) + 5

h(x) = infinity

Therefore, as x  approaches the vertical asymptote,  [tex]h(x)[/tex] →[tex]\infty[/tex]  and as x approaches to positive [tex]\infty[/tex], h(x) →[tex]-\infty[/tex]

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The area of the surface obtained by rotating the curve y = (x³/6) + (1/2x) from 1/2 to 1 about the x-axis is given by the above expression is 2π (1/6 x √(1+ 9x² - 3x⁴/4)).

Calculate the arc length of the curve

We first need to calculate the arc length of the curve, which can be done using the formula:

L = ∫aᵇ √(1+ (dy/dx)²) dx

where,

dy/dx = (3x² - 1/2x²)/6

Therefore, the arc length of the curve is given by:

L = ∫1/2¹√(1+ (3x² - 1/2x² )/6)dx

Calculate the area of the surface

Once we have the arc length of the curve, we can calculate the area of the surface obtained by rotating the curve about the x-axis. This can be done using the formula:

A = 2π × L

Substituting the arc length of the curve in the formula, we get:

A = 2π × ∫1/2¹√(1+ (3x² - 1/2x²)/6)dx

Evaluate the integral

Finally, we need to evaluate the integral in order to calculate the area of the surface. We can do this using integration by parts, which gives us:

A = 2π × ∫1/2¹√(1+ (3x² - 1/2x²)/6)dx

= 2π (1/6 x √(1+ 9x² - 3x⁴/4) - (1/6) ∫1/2¹ (9x² - 3x⁴/4)/√(1+ 9x² - 3x⁴4) dx)

Therefore, the area of  the surface is 2π (1/6 x √(1+ 9x² - 3x⁴/4)).

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The median is the best measure of center and equals 11 from box plot

A box plot uses a number line from 6 to 21 with tick marks every one-half unit.

The box extends from 10 to 15 on the number line.

A line in the box is at 11. The lines outside the box end at 7 and 20.

Based on the information provided in the box plot, the best measure of center for the data shown is the median.

The median is represented by the line within the box, which is at 11. Therefore, the best measure of center for the data is the median, and its value is 11.

Hence, the median is the best measure of center and equals 11.

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Answer:

129*

Step-by-step explanation:

First notice the 2 parallel lines. Since the 2 parallel lines are intersecting the same line, the angle measures would be the same for both. Since the first angle is 51*, the corresponding angle for the next line is also 51*. Since it intersects a straight line, the angles need to add up to 180*. 51 + x = 180 so x = 129*

Por lo tanto, el valor de 72 + 68 - 59 es 81.

Para encontrar el valor de la expresión 72 + 68 - 59, primero necesitamos determinar el valor de "a" en la ecuación dada.

Dado que "ax98" es un número impar y "ax99" también es impar, podemos concluir que "a" debe ser un número impar. Supongamos que "a" es igual a algún número impar "x".

Ahora podemos sustituir el valor de "a" en la expresión 72 + 68 - 59:

72 + 68 - 59 = 140 - 59 = 81

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The solution of the two system of equations using elimination method is x = 8, and y = 3.

The solution of the two system of equations using elimination method is calculated as follows;

The given equations;

2x - 5y = 1   -------- (1)

-3x + 2y = -18 ----- (2)

To eliminate x, multiply equation (1)  by 3 and equation (2) by 2, and add the to equations together;

3:   6x  -  15y = 3

2:   -6x  + 4y  = -36

-----------------------------------

             -11y = -33

               y = 33/11  =  3

Now, solve for the value of x by substituting the value of y back into any of the equations.

2x - 5y = 1

2x - 5(3) = 1

2x - 15 = 1

2x = 16

x = 8

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To solve the given initial-value problem, we can separate the variables and integrate both sides. The final answer is Therefore, the solution to the initial-value problem is y = [tex]\frac{-1}{(-14t - 1)}[/tex], where y(0) = 1.

Given: y' = [tex]-(14)y^2\\[/tex]

Initial condition: y(0) = 1

Separating variables:

[tex]\frac{Dy}{y^2 }[/tex]= -14 dt

Integrating both sides:

∫([tex]\frac{1}{y^2}[/tex]) dy = ∫-14 dt

Integrating the left side:

[tex]\frac{-1}{y}[/tex]= -14t + C1

Solving for y:

y = [tex]\frac{-1}{-14t + C1}[/tex]

Using the initial condition y(0) = 1:

1 = [tex]\frac{-1}{C1}[/tex]

C1 = -1

Substituting the value of C1 back into the solution:

y = [tex]\frac{-1}{(-14t - 1)}[/tex]

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The volume of the square pyramid that has sides of length 15 cm and height of 14 cm is: D. 1050 cm³

To find the volume of a square pyramid, you can use the formula: Volume = (1/3) * Base Area * Height.

Since the base of the square pyramid has sides of length 15 cm, the base area can be calculated as:

Base area = 15 cm * 15 cm

= 225 cm².

Plugging the values into the formula, the volume of the pyramid:

= (1/3) * 225 * 14 cm

= 1050 cm³.

Therefore, the volume of the square pyramid is 1050 cm³.

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Answer:

1.5.1 : constant difference is 8

1.5.2: When there are 10 chairs stacked, it's 127 cm tall.

Step-by-step explanation:

There is a linear relationship between the height of the stack and the number of chairs.

1 chair = 55 cm + 0 extra cm = 55cm

2 chairs = 55cm + 8cm = 63 cm

3 chairs = 55cm + 8(2)cm = 71 cm

4 chairs = 55cm + 8(3)cm = 79 cm

1.5.1 the constant difference between all the underlined numbers above is 8.

1.5.2 You could just keep calculating above until you get 127 cm. (Your teacher might not like that, but it's an option!)

You can find the equation & either solve for the number of chairs OR graph it.

So if we let C = the number of stacked chairs, our equation for H (height) would be:

H = 55 + 8(C-1)

If we substitute H = 127, solve for C.

127 = 55+ 8(c-1)

127 = 55+ 8c-8

127 = 47 + 8c

127 -47 = 8c

80 = 8c

10=c

When there are 10 chairs stacked, it's 127 cm tall.

Check that the answer works:

55 cm (1st chair) + 8*9 (8cm for each additional chair) = 55+ 72 = 127 cm

The distance from my town to the little village is $35$ miles.

To find the distance from my town to the little village, we need to add up the distances of each segment of the trip. We know that the straight-line distance from the capital city to the little village is $140$ miles, but we can't use that information directly. Instead, we need to use the distances between each town.
From the capital city to my town is $80$ miles, from my town to your town is $25$ miles, and from your town to the little village is $35$ miles. Adding those distances gives us:
$80 + 25 + 35 = 140$ miles
So the distance from my town to the little village is $35$ miles.

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An ordered field is a set F with two binary operations, addition (+) and multiplication (⋅), and a binary relation, the order relation (≤), that satisfy certain axioms.

In an ordered field, we have the following properties: commutativity, associativity, distributivity, identity, inverses, transitivity, totality, antisymmetry, and the order-preserving nature of addition and multiplication.
To prove that every ordered field has no smallest positive element, we use a proof by contradiction. Suppose there exists an ordered field F and a smallest positive element ε > 0 in F. In an ordered field, the product of two positive elements is positive, and since ε is the smallest positive element, we must have ε² > ε.
However, due to the properties of ordered fields, we can perform the following manipulations: ε² > ε implies ε² - ε > 0, which in turn implies ε(ε - 1) > 0. Since ε is positive, we can conclude that ε - 1 must also be positive. Therefore, ε > 1.
But now, we have another positive element, 1, which is smaller than ε, contradicting our assumption that ε is the smallest positive element in the ordered field. This contradiction proves that no ordered field can have a smallest positive element, as any potential candidate would lead to the discovery of an even smaller positive element.

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Answer:

-2/7 is 3 and -6 is 2

Step-by-step explanation:

to make a slope perpendicular, you need to swap the numerator and denominator and multiply by negative so - becomes + and + becomes -

Answer:

-2/7 is 3 and -6 is 2

Step-by-step explanation:

Answer:

Line n

Step-by-step explanation:

Line n, the pink line, it's just y=n, it doesn't change no matter what the x is.

The owner of the bookstore sells the used books for $6 each. J.

The price of a used book in the bookstore we need to calculate how much the owner is selling the books for.

The owner of the bookstore buys the used books from customers for $1.50 each.

The owner resells the used books for we need to multiply the cost price by 400%:

$1.50 x 400% = $1.50 x 4

= $6

The markup percentage for the used books is very high.

The owner is reselling the used books for four times the amount he paid for them.

This is a common practice in the used book industry as it allows the owner to make a profit on the books they sell.

It is important for customers to be aware of the markup and shop around for the best prices.

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Answer:

should be 18 ft

Step-by-step explanation:

6÷4= 1.5

27÷1.5= 18

Answer:

To begin, we set each component to zero:

x + 3 = 0 x^2 - 4 = 0

When we solve for x in the first equation, we get:

x = -3

We may factor the second equation further using the difference of squares formula which is:

(x + 2)(x - 2) = 0

Then, in each factor, we solve for x:

x + 2 = 0 or x - 2 = 0

x = -2 or x = 2

As a result, the function's zeroes are x = -3, x = -2, and x = 2.

Answer:

-3, -2, and 2.

Step-by-step explanation:

Solving for x in the first equation gives:

x+3 = 0

x = -3

Solving for x in the second equation gives:

x^2-4 = 0

(x+2)(x-2) = 0

x+2 = 0 or x-2 = 0

x = -2 or x = 2

Therefore, the zeroes of the function F(x) are -3, -2, and 2.

By algebra properties, the simplified form of the expressions are listed below in the following four cases:

Case 1: 6

Case 2: 1 / 5

Case 3: √3

Case 4: - 3

In this problem we must simplify expressions involving powers and roots by algebra properties, mainly power and root properties. Now we proceed to show how each expression is simplified:

Case 1

[tex](1^{3}+2^{3}+ 3^{3})^{\frac {1}{2}}[/tex]

[tex](1 + 8 + 27)^{\frac{1}{2}}[/tex]

[tex]36^{\frac{1}{2}}[/tex]

√36

6

Case 2

[tex]\left[\left(625^{-\frac{1}{2}}\right)^{-\frac{1}{4}}\right]^{2}[/tex]

[tex]\left[\left[\left(625^{\frac{1}{2}}\right)^{-1}\right]^{-\frac {1}{4}}\right]^{2}[/tex]

[tex]\left[\left(\frac{1}{25}\right)^{\frac{1}{4}}\right]^{2}[/tex]

[tex]\left(\frac{1}{25} \right)^{\frac{1}{2}}[/tex]

√(1 / 25)

1 / 5

Case 3

[tex]\frac{9^{\frac{1}{2}}\times 27^{- \frac {1}{3}}}{3^{\frac{1}{6}}\times 3^{-\frac{2}{3}}}[/tex]

[tex]\frac{(3^{2})^{\frac{1}{2}}\times (3^{3})^{-\frac{1}{3}}}{3^{\frac{1}{6}}\times 3^{-\frac{2}{3}}}[/tex]

[tex]\frac{3\times 3^{- 1}}{3^{\frac{1}{6}}\times 3^{- \frac{2}{3}}}\\[/tex]

[tex]\frac{1}{3^{-\frac{1}{2}}}[/tex]

[tex]3^{\frac{1}{2}}[/tex]

√3

Case 4

[tex]64^{-\frac{1}{3}}\cdot \left[64^{\frac{1}{3}}-64^{\frac{2}{3}}\right][/tex]

[tex]64^{-\frac{1}{3}}\cdot 64^{\frac{1}{3}}-64^{-\frac{1}{3}}\cdot 64^{\frac{2}{3}}[/tex]

[tex]1 - 64^{ \frac{1}{3}}[/tex]

1 - ∛64

1 - 4

- 3

To learn more on powers and roots: https://brainly.com/question/2165759

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