Let A be a set With 12 elements. (a) Find the number of subsets of A. (b) Find the number of subsets of A having one or more elements. (c) Find the number of subsets of A having exactly one element. (d) Find the number of subsets of A having two or more elements. [Hint: Use the answers to parts (b) and (c).]

The polynomial function modeled by the given diagram is given as follows:

p(x) = 3x² - 7x - 6.

The polynomial function modeled by the given diagram is obtained considering the keys of the problem, which are the terms represented by each figure.

The polynomial is constructed as follows:

Then the expression used to construct the polynomial is given as follows:

p(x) = 3x² + 2x - 9x - 6.

Combining the like terms, the polynomial function is defined as follows:

p(x) = 3x² - 7x - 6.

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91 student tickets were sold and 37 adults tickets were sold whose total 128 tickets were sold.

The elimination method is a technique for solving a system of linear equations, which involves adding or subtracting the equations to eliminate one of the variables, and then solving for the other variable.

Let x be the number of student tickets sold, and y be the number of adult tickets sold. Then we can set up a system of two equations to represent the information given:

x + y = 128 (1) (the total number of tickets sold is 128)

5x + 8y = 751 (2) (the total revenue from ticket sales is $751)

We can solve for one of the variables in terms of the other in the first equation:

x = 128 - y

Substituting this expression into the second equation to eliminate x, we get:

5(128 - y) + 8y = 751

Expanding and simplifying:

640 - 5y + 8y = 751

3y = 111

y = 37

Therefore, 37 adult tickets were sold. Substituting this value back into equation (1) to solve for x, we get:

x + 37 = 128

x = 91

Therefore, 91 student tickets were sold.

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The arc length of the sector is 7.85 meters whose circumference is 20π meters.

In geometry, a sector is a part of a circle enclosed by two radii and an arc. The arc of a sector is a portion of the circumference of the circle. Sectors are often used in geometry and trigonometry to calculate areas, angles, and other measurements.

The equation for a circle's circumference is:

C = 2πr

Since we know the circumference of the circle, we can solve for the radius:

20π = 2πr

Dividing both sides by 2π, we get:

r = 10

Arc length = (central angle / 360°) x 2πr

where r is the radius and the central angle is in degrees.

Plugging in the given values, we get:

Arc length = (45° / 360°) x 2 x 3.14 x 10

Arc length = (1/8) x 62.8

Arc length = 7.85 meters

Therefore, the arc length of the sector is 7.85 meters.

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Answer:

The answer is Nitrogen

Hope this helps :)

Answer:

Yes, No, No.

Explanation:

Answer:

Place an X for the first box as [Yes], [No], [No]

Step-by-step explanation:

When we enter x=2 and y=1 into the first system of equations, we can see that both conditions are met. Thus the answer to this system is (2, 1).

When x=2 and y=1 are substituted into the second system of equations, we obtain 1=-3 and 1=-2, which are false, and so (2, 1) is not a solution for this system.

When x=2 and y=1 are substituted into the third system of equations, the results are -3=-2 and 1=-3, which are false, hence (2, 1) is not a solution for this system.

If a customer purchases 3 of the products, the probability of getting at least one that is defective is 38.59%.

In order to determine the probability of getting at least one defective product if a customer purchases three products with a defective rate of 7.5%, we can use the concept of complementary probability.

The probability of getting at least one defective product can be calculated as the complement of the probability of getting none defective products.

So, the probability of getting no defective products is:

P(none defective) = (1 - 0.075)³ = 0.6141

Therefore, the probability of getting at least one defective product is:

P(at least one defective) = 1 - P(none defective) = 1 - 0.6141 = 0.3859 or 38.59%

.So, the probability of getting at least one that is defective is 38.59%.

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A) Based on the scatter plot, an r value of 0.98 seems to be a reasonable estimate of the correlation between the number of flowers and the days in March. The scatter plot shows an upward trend with points that are moderately spread out from a line of best fit.

B) A scenario that would be a causal relationship for flowers in a garden could be the amount of sunlight the garden receives.

A) Based on the scatter plot, an r value of 0.98 seems to be a reasonable estimate of the correlation between the number of flowers and the days in March. The scatter plot shows an upward trend with points that are moderately spread out from a line of best fit. This indicates that there is a strong positive relationship between the number of flowers and the days in March, which is reflected in the high correlation coefficient. Therefore, it is likely that the r value of 0.98 is an accurate value for this data.

B) A scenario that would be a causal relationship for flowers in a garden could be the amount of sunlight the garden receives. For example, if the garden receives more sunlight, it could cause the flowers to grow more quickly and bloom earlier in the month. On the other hand, if the garden receives less sunlight, the flowers may take longer to grow and bloom, and there may be fewer flowers overall. In this scenario, sunlight would be the independent variable, and the number of flowers bloomed would be the dependent variable.

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The given question is incomplete, the complete question is:

He scatter plot shows the number of flowers that have bloomed in the garden during the month of March:

A scatter plot with points showing an upward trend with points that are moderately spread out from a line of best fit. The y axis is labeled Number of Flowers and the x axis is labeled Days in March

Part A: Using computer software, a correlation coefficient of r = 0. 98 was calculated. Based on the scatter plot, is that an accurate value for this data? Why or why not? (5 points)

Part B: Instead of comparing the number of flowers and the day in March, write a scenario that would be a causal relationship for flowers in a garden. (5 points)

Answer:

y = 1

Step-by-step explanation:

3x + 4y = 34            x = 10

3(10) + 4y = 34

30 + 4y = 34

4y = 4

y = 1

So, the y-value is 1 when x = 10

[tex]\huge\text{Hey there!}[/tex]

[tex]\mathsf{3x + 4y = 34}[/tex]

[tex]\mathsf{3(10) + 4y = 34}[/tex]

[tex]\mathsf{30 + 4y = 34}[/tex]

[tex]\mathsf{4y + 30 = 34}[/tex]

[tex]\large\text{SUBTRACT 30 to BOTH SIDES}[/tex]

[tex]\mathsf{4y + 30 - 30 = 34 - 30}[/tex]

[tex]\large\text{SIMPLIFY IT!}[/tex]
[tex]\mathsf{4y = 34 - 30}[/tex]

[tex]\mathsf{4y = 4 }[/tex]

[tex]\large\text{DIVIDE 4 to BOTH SIDES}[/tex]

[tex]\mathsf{\dfrac{4y}{4} = \dfrac{4}{4}}[/tex]

[tex]\large\text{SIMPLIFY IT!}[/tex]

[tex]\mathsf{y = \dfrac{4}{4}}[/tex]

[tex]\mathsf{y = 1}[/tex]

[tex]\huge\text{Therefore your answer should most likely be:}[/tex]
[tex]\huge\boxed{\mathsf{y = 1}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]

Answer:(a + 1) (3a + x)

Step-by-step explanation:

Factor a+1 out of  3a ( a + 1 ) + x (a + 1 )

Hope this helps

In tests of significance about an unknown parameter of some population, "We observe a value of an estimate of the unknown parameter based on a simple random sample from the population that is very unlikely to occur if the null hypothesis is true" is considered strong evidence against the null hypothesis. The correct answer is Option (D).

We apply the principle of hypothesis testing to test a population's claims in inferential statistics. The null hypothesis (H₀) is always a statement about the population parameter that we believe to be true. However, we use the sample data to decide whether the null hypothesis is true or not. When we perform the hypothesis testing, we must consider the level of significance, the sample size, and the nature of the test.

The value of an estimate of the unknown parameter based on a simple random sample from the population that is very unlikely to occur if the null hypothesis is true is considered strong evidence against the null hypothesis. In other words, if the value of the test statistic is greater than the critical value, we can reject the null hypothesis. Consequently, we will have sufficient evidence to support the alternative hypothesis.

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Answer:

Step-by-step explanation:

F = (-1,9)

G = (-3,11)

the initial speed with which the particle was projected is approximately 29.7 m/s.

How to solve?

We can solve this problem using the equations of motion for a particle moving under constant acceleration due to gravity.

Let v be the initial velocity with which the particle was projected, and let θ be the angle of projection with respect to the horizontal. Since the particle is launched from a height of 15m, its initial vertical velocity is v sin(θ), and its initial horizontal velocity is v cos(θ).

The time taken for the particle to hit the ground can be found by using the equation:

y = y0 + v0t + 1/2at²

where y is the vertical displacement, y0 is the initial vertical position, v0 is the initial vertical velocity, a is the acceleration due to gravity (-9.8 m/s²), and t is the time taken.

Since the particle starts and ends at the same vertical position (15m), we have:

y - y0 = 0

Substituting the values, we get:

0 = 15 + v sin(θ)t - 1/2(9.8)t²

Simplifying and rearranging, we get:

4.9t² - vt - 30 = 0

Using the quadratic formula, we get:

t = (v ±√(v² + 4(4.9)(30))) / (2(4.9))

Since we want the time taken for the particle to hit the ground, we take the positive root:

t = (v + √(v² + 588)) / 9.8

Now, we can use the horizontal displacement to find the initial speed v. Since the particle travels a horizontal distance of 30m in time t, we have:

x = v cos(θ) t

Substituting the values, we get:

30 = v cos(θ) [(v +√(v² + 588)) / 9.8]

Simplifying and rearranging, we get:

v² - 882 = 0

Using the quadratic formula again, we get:

v = ±√(882)

Since the initial velocity must be positive, we take the positive root:

v ≈ 29.7 m/s

Therefore, the initial speed with which the particle was projected is approximately 29.7 m/s.

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The scientist should group the data into categories or bins of 0.5 millimeters, such as 1.0-1.5, 1.5-2.0, and 2.0-2.5,

To create a histogram to show the distribution of growth across the population, the scientist needs to group the data into categories or bins. The size of the bins will determine the shape of the histogram and how well it represents the data.

One way to determine the bin size is to calculate the range of the data and divide it by the number of bins desired. Another approach is to use a standard bin size, such as 0.5 or 1.0.

For this particular data set, a reasonable bin size could be 0.5 millimeters. The data can then be grouped into the following bins:

1.0 - 1.5

1.5 - 2.0

2.0 - 2.5

This will result in three bins that cover the entire range of the data. The scientist can then count the number of insects that fall into each bin and create a histogram to show the distribution of growth across the population.

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Answer:

-21

Step-by-step explanation:

y^2 + 8y - 9                       y = -6

(-6)² + 8(-6) - 9

36 - 48 - 9

-21

So, the answer is -21

Answer:y=\frac{7}{12}-i\frac{\sqrt{167}}{12},\:y=\frac{7}{12}+i\frac{\sqrt{167}}{12}

Step-by-step explanation:y=\frac{7}{12}-i\frac{\sqrt{167}}{12},\:y=\frac{7}{12}+i\frac{\sqrt{167}}{12}

The matrix of predicted rankings UVT is [[48,24,12],[0,0,0],[24,12,6]].

The matrix of predicted rankings UVT can be calculated using the formula UVT = UV.The provided initial values for U() and V(0) are as follows:U() = [[2,1],[1,0],[3,-1]]V(0) = [[24,12,6],[0,0,0],[12,6,3]]Using the above values, the matrix of predicted rankings UVT can be computed as follows:UVT = UVU = [[2,1],[1,0],[3,-1]]V = [[24,12,6],[0,0,0],[12,6,3]]UVT = [[2,1],[1,0],[3,-1]] x [[24,12,6],[0,0,0],[12,6,3]]= [[48,24,12],[0,0,0],[24,12,6]]Therefore, the matrix of predicted rankings UVT is [[48,24,12],[0,0,0],[24,12,6]].

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Considering flow over a flat plate, and by using the Thwaites-Walz method and the Pohlhausen method are very similar, but they differ significantly from the exact solution.

The Thwaites-Walz Method for flow over a flat plate:

          The Blasius method can be used to obtain the non-dimensional velocity distribution over a flat plate. But the computation of the shear stress and friction coefficient from this velocity distribution requires the knowledge of the second derivative of u with respect to y which is difficult to obtain.

          The Thwaites method is an alternative method for computing the friction coefficient, which avoids the computation of the second derivative of u with respect to y. This method involves the solution of an ordinary differential equation.

            This method is particularly useful for computing the friction coefficient in the early stages of the boundary layer. The equations for the Thwaites method are as follows:

                 [tex]\frac{d^2\delta}{dx^2} =\frac{\delta}{u^2}\left(1+ \frac{\delta}{2}\frac{dU/dx}{U}\right)C_f[/tex]

                                                         = [tex]\frac{0.288\delta}{Re_x}(\frac{d\delta}{dx})^{1/2}Re_x[/tex]

                                                         = [tex]\frac{\rho u(x)x}{\mu}\tau_w[/tex]

                                                         = [tex]\rho u_\infty C_f/2x[/tex]

                                                         = [tex]\frac{1}{C_f}\int_{0}^{\delta}u_\infty \left(1- \frac{u}{u_\infty}\right)dy$$[/tex]

The following are the predictions using the Thwaites-Walz method to predict d, d*, 8, and

                   [tex]Cvs x.*d = 0.375 x^(1/5)*d*[/tex]

                                    = [tex]4.91 x^(1/5)*8[/tex]

                                    = [tex]0.664 x^(3/5)*Cv[/tex]

                                    = [tex]1.328 x^(1/5)[/tex]

        The Pohlhausen method is a simple method for computing the shear stress and the friction coefficient, which is based on an approximate solution of the boundary layer equations. The Pohlhausen method is based on the assumption that the velocity distribution is a parabolic function of the distance from the wall.

             The equations for the Pohlhausen method are as follows:

                      [tex]u(x,y)= U(x)\left(1-\left(\frac{y}{\delta}\right)^2\right)\tau_w[/tex]

                                 = [tex]\rho u_\infty \frac{dU}{dx}\frac{\delta^2}{3}C_f[/tex]

                                 =  [tex]\frac{2}{3}\frac{\tau_w}{\rho u_\infty^2}x[/tex]

                                 =  [tex]\frac{1}{C_f}\int_{0}^{\delta}u_\infty \left(1- \frac{u}{u_\infty}\right)dy$$[/tex]

The following are the predictions using the Pohlhausen method to predict d, d*, 8, and

                       Cvs x.• d = 0.37 x^(1/5)• d*

                                       = 4.9 x^(1/5)• 8

                                       = 0.664 x^(3/5)• Cv

                                       = 1.328 x^(1/5)

The following are the exact solutions for flow over a flat plate. Equations (2.21) and (2.22) are for the shear stress and friction coefficient respectively.

                           [tex]$$ \tau_w = \rho u_\infty C_f/2[/tex]

                                =  [tex]\frac{0.664 \rho u_\infty^2 x^{3/5}}{Re_x^{1/5}}C_f[/tex]

                                 =   [tex]\frac{0.664}{Re_x^{1/2}}[/tex]

          The following are the predictions using the exact solutions for flow over a flat plate.

                                  [tex]*d = 0.664 x^(3/10)*d*[/tex]

                                        = [tex]4.91 x^(1/5)*8[/tex]

                                       = [tex]0.664 x^(3/5)*Cv[/tex]

                                       = [tex]1.328 x^(1/5)[/tex]

         Hence, the predictions using the Thwaites-Walz method and the Pohlhausen method are very similar, but they differ significantly from the exact solution.

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The answer is: A) The data support the researchers' belief that night owls have better dream recall than early birds. B) we can be 92% confident that the true difference in mean number of dreams recalled per week between night owls and early birds is between 1.87 and 2.63.

A) These are the alternative and null hypotheses:

H0: μE = μN (the mean number of dreams recalled each week is the same for early birds and night owls) (the mean number of dreams recalled per week is the same for early birds and night owls)

Ha: μE < μN (the mean number of dreams recalled each week is smaller for early birds than for night owls) (the mean number of dreams recalled per week is lower for early birds than for night owls)

Using the following formula, we can run a two-sample t-test with unequal variances:

t = [(sN2 / nN) + (sE2 / nE)] / sqrt[(xN - xE)]

where nN and nE are the sample sizes for night owls and early birds, respectively, and xN and sN and xE and sE are the sample means and standard deviations for night owls and early birds, respectively.

When we enter the values, we obtain:

t = (9.55 - 7.26) / sqrt[(5.88^2 / 100) + (6.94^2 / 100)] = 5.01

The data are consistent with the researchers' hypothesis that night owls are more capable of remembering their dreams than early birds.

B) We can use the following formula to determine the confidence interval:

CI is equal to (xN - xE) t/2 * sqrt[(sN / nN) + (sE / nE)].

where t/2 is the t-value for the required level of confidence and degrees of freedom, and xN, xE, sN, sE, nN, and nE are the same as previously (198 in this case).

With a t-value of 1.75 and a 92% confidence level (from a t-distribution with 198 degrees of freedom), we get:

CI is equal to (9.55 - 7.26) 1.75 * sqrt[(5.88 - + 6.94 / 100)] = (1.87, 2.63) (1.87, 2.63)

The genuine difference between night owls and early birds in terms of the average number of dreams recalled per week is therefore between 1.87 and 2.63, with a 92% confidence interval.

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Correct Option is Shifted 2 units left and 4 units up

A triangle is a geometric shape that is formed by three straight line segments that connect three non-collinear points. The three points where the segments intersect are called the vertices of the triangle, while the segments themselves are called the sides. The area enclosed by the sides of the triangle is called its interior, while the space outside the triangle is called its exterior.

The vertices of ABC are (4, 6), (7, 6), and (5,9)

The transformed image A'B'C' has vertices as

(2,10) (5,10) (3,13)

We see a pattern when we compare the matching vertices.

The y coordinate is raised by 4, while the x coordinate is shrunk by 2.

This implies the transformation is

Shifted 2 units left and 4 units up

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Answer:

Shifted 2 units left and 4 units up

Step-by-step explanation:

hope this helps

The equation that best models the given graph is given by `y = -100x + 2200` where `y` represents the expected monthly revenue and `x` represents the price per roll.

The equation that best models the given graph of the expected monthly revenue from rolls of chicken wire, based on the price per roll, is given by `y = -100x + 2200` where `y` represents the expected monthly revenue and `x` represents the price per roll.Step-by-step explanation:The graph given below shows the expected monthly revenue from rolls of chicken wire, based on the price per roll.From the graph, we can see that as the price per roll increases, the expected monthly revenue decreases.

So, the equation that models this situation should have a negative slope.Now, let's find the slope of the line passing through the points `(20, 1200)` and `(0, 2200)` using the slope formula. The slope formula is given by:$$\text{slope} = \frac{y_2 - y_1}{x_2 - x_1}$$Here, we have `x_1 = 20`, `y_1 = 1200`, `x_2 = 0`, and `y_2 = 2200`. So, substituting the values, we get:$$\text{slope} = \frac{2200 - 1200}{0 - 20}$$$$\text{slope} = -\frac{1000}{20}$$$$\text{slope} = -50$$So, the equation of the line is of the form:$$y = mx + b$$where `m` is the slope and `b` is the y-intercept.From the graph, we can see that the y-intercept is `2200`.

So, substituting the values of `m` and `b` in the above equation, we get:$$y = -50x + 2200$$

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Answer:

160º

Step-by-step explanation:

Length of an arc = 2πr(θ/360º)

40π/3 = 2πr(θ/360)

20 = (3x15)(θ/360)

20 x 360 = 45θ

θ = 7200/45 = 160º

Answer:

160⁰

Step-by-step explanation:

all is included in the picture, just use the formula and substitute the values

Answer:

Step-by-step explanation:

2x - 3 = 6 + x

2x - x = 6 + 3

x = 9

    2.        2(2x-3)=6+x

2(2x - 3) = 6 + x

4x - 6 = 6 + x

4x - x = 6 + 6

3x = 12

x = 12 : 3

x = 4

The significance of a pedigree symbol consisting of a square with a diagonal slash mark through it is that it represents the male members of the family

It is a standard symbol in a pedigree chart. This symbol represents the sex of a person, in this case, it represents the male sex. In other words, a square with a diagonal slash mark through it is used to represent the male gender in pedigree charts.

A pedigree chart is a diagram that shows the genetic relationships between individuals in a family. It is used to track genetic diseases or traits through generations. A pedigree chart can provide information about a family's medical history and can help doctors to understand how genetic disorders are inherited from one generation to the next. Each symbol used in the pedigree chart has a specific meaning.

The squares represent males while the circles represent females. The horizontal line between two symbols represents marriage, and the vertical line from a symbol represents a child of that union.

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The average number of beads that the girls have is 63

Let's start by using algebra to represent the given information:

Let b be the number of beads that Sally has.

Then, Kelly has 3/4 times as many beads as Sally, which can be expressed as (3/4)b.

Also, we know that Kelly has 18 more beads than Sally, which can be expressed as (b + 18).

Putting these together, we can write the equation:

(3/4)b = b + 18

Solving for b, we get:

b = 72

So, Sally has 72 beads, and Kelly has (3/4) × 72 = 54 beads.

The average number of beads that the girls have is (72 + 54)/2 = 63 beads

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The critical points of function f(x,y) are (0,0) and (2,1), and (2,1) is a local maximum.

To find the critical points of f(x,y), we need to find all values of (x,y) where the gradient of f(x,y) equals zero. The gradient of f(x,y) is given by:

∇f(x,y) = <(y-2xy), (x-2y^2)>

Setting each component of the gradient equal to zero yields two equations:

y - 2xy = 0

x - 2y^2 = 0

Solving these equations simultaneously, we obtain two critical points: (0,0) and (2,1).

To determine the nature of each critical point, we compute the Hessian matrix of f(x,y):

H(f) = [ 2y -2x ]

[-2y 4y ]

At (0,0), H(f) = [0 0; 0 0], which is a degenerate matrix. Therefore, we cannot use the second derivative test to determine the nature of this critical point.

At (2,1), H(f) = [2 -4; -2 4], which has a negative determinant and a positive trace. Therefore, by the second derivative test, we conclude that (2,1) is a local maximum.

In summary, the critical points of f(x,y) are (0,0) and (2,1), and (2,1) is a local maximum.

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Answer:

x = 4[tex]\sqrt{2}[/tex]

Step-by-step explanation:

using the cosine ratio in the lower right triangle and the exact value

cos45° = [tex]\frac{1}{\sqrt{2} } }[/tex] , then

cos45° = [tex]\frac{adjacent}{hypotenuse}[/tex] = [tex]\frac{4}{x}[/tex] = [tex]\frac{1}{\sqrt{2} }[/tex]  ( cross- multiply )

x = 4[tex]\sqrt{2}[/tex]

The equation of motion of the system is, `x(t) = Acos⁡(ωt + ϕ)` where `ω = √(k/m)` is the angular frequency of the system, `A` is the amplitude of motion, `ϕ` is the phase angle, `k` is the spring constant, and `m` is the mass attached to the spring. The period of simple harmonic motion (in seconds) is 0.628` seconds (approx).

The mass weighing 16 pounds is attached to a spring whose spring constant is 25 lb/ft.

So, the mass of the system `m = 16/32 = 0.5` slugs (1 slug = 32 lb.s^2/ft).

Thus, the angular frequency of the system is, `ω = √(k/m) = √(25/0.5) = 10` rad/s.

So, the equation of motion of the system is,x(t) = Acos⁡(10t + ϕ)

Given that, x(0) = 16/25, x(t) = Acos⁡(10t + ϕ) ...(1)

At t = 0, x(0) = Acos⁡ϕ = 16/25

So, `A = (16/25)/cos⁡ϕ`.

Therefore, by substituting `A` in equation (1), we get

x(t) = (16/25)/cos⁡ϕ × cos⁡(10t + ϕ) = 0.64 cos⁡(10t + ϕ)/cos⁡ϕ

Comparing this equation with the given equation, x(t) = -16 cos⁡(251), we get`10t + ϕ = 251`, `cos⁡ϕ = -16/25`

Therefore, `ϕ = cos^{-1}⁡(-16/25) = 123.7°`.The period of simple harmonic motion (in seconds) is given by,

`T = 2π/ω = 2π/10 = 0.628` seconds (approx).

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For the given ratio of ingredients 3/14 of Alex's drink is made from blackcurrants.

Comparing two numbers or values using ratios, which are often stated as fractions or colons. In mathematics, ratios are used to represent connections between various objects or numbers, such as the ratio of a rectangle's length to breadth or the proportion of boys to girls in a class. Ratios can be sped up or stated in a variety of ways, such decimals or percentages.

The given ratio for blackcurrants to raspberries is 2:5.

The total ratio is:

2 + 5 = 7

Hence, the fraction of the puree made from blackcurrants is = 2/7.

Now, ratio of fruit puree to milk is 3:1 = 3 + 1 = 4.

Hence, the fraction of the milkshake made from fruit puree is = 3/4.

For blackcurrants we have:

(2/7) * (3/4) = 6/28 = 3/14

Hence, for the given ratio of ingredients 3/14 of Alex's drink is made from blackcurrants.

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Answer:

Step-by-step explanation:

Rewrite

3

a

2

10

b

6

as

(

a

b

3

)

2

3

10

.

√

(

a

b

3

)

2

3

10

Pull terms out from under the radical.

a

b

3

√

3

10

Rewrite

√

3

10

as

√

3

√

10

.

a

b

3

⋅

√

3

√

10

Combine.

a

√

3

b

3

√

10

Multiply

a

√

3

b

3

√

10

by

√

10

√

10

.

a

√

3

b

3

√

10

⋅

√

10

√

10

Combine and simplify the denominator.

√

3

√

10

b

3

⋅

10

Simplify the numerator.

Tap for more steps...

a

√

30

b

3

⋅

10

Move

10

to the left of

b

3

.

a

√

30

10

b

3

Step-by-step explanation:

[tex]{ \tt{ \sqrt{ \frac{3 {a}^{2} }{10 {b}^{6} } } }} = { \tt{ \frac{ {(3a {}^{2}) }^{ \frac{1}{2} } }{ {(10b {}^{6} )}^{ \frac{1}{2} } } }} \\ \\ = { \tt{ \frac{a \sqrt{3} }{ {b}^{3} \sqrt{10} } }} \\ \\ = { \tt{ \frac{a \sqrt{3} }{ {b}^{3} \sqrt{10} } }}[/tex]