Two containers designed to hold water are side by side, both in the shape of a cylinder. Container A has a diameter of 14 feet and a height of 11 feet. Container B has a diameter of 10 feet and a height of 19 feet. Container A is full of water and the water is pumped into Container B until Container B is completely full.

The probability of drawing one card from a standard deck of cards and choosing a "face card" is 3/13.

What is probability?

Probability is a measure of the likelihood or chance of an event occurring, expressed as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.

There are a total of 12 face cards in a standard deck of 52 cards (4 jacks, 4 queens, and 4 kings).

The probability of drawing a face card can be calculated by dividing the number of face cards by the total number of cards in the deck:

P(face card) = number of face cards / total number of cards

[tex]P(face \: card) = \frac{12}{52} \\ P(face \: card) = \frac{3}{13} [/tex]

Therefore, the probability is D. 3/13.

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The shape that would be seen if you were to take a cross-section parallel to the base of a rectangular prism is a RECTANGLE.

Here, we have,

we know that,

A regular prism is a base with a regular polygon, whereas a prism whose base is an irregular polygon is called an irregular prism.

so, we have,

This is because the cross-section of a rectangular prism will always be a rectangle.

The definition of a prism is that the cross-section parallel to the base will be uniform.

Hence, The shape that would be seen if you were to take a cross-section parallel to the base of a rectangular prism is a RECTANGLE.

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The measure of m(ZR) would be 145°.

In geometry, a secant is a straight line that intersects a curve at two or more points. A secant line is used to study the properties of the curve such as its slope, curvature, and points of intersection with other curves. In the context of circles, a secant is a line that intersects a circle at two points, creating a chord. A secant is different from a tangent, which is a line that intersects a curve or circle at only one point and is perpendicular to the curve at that point.

Now we know that if Two secants intersect inside the circle.

Then according a property of intersecting chords.

m(ZR) - m(KV) = 2 (30°)

(5x+10)° - (3x+4)°  = 60°

(5x+10° - 3x - 4°)= 60°

2x+6°= 60°

2x = 60° - 6°

2x = 54°

x = 27°

Now put the value in m(ZR)

m(ZR) = (5x27+10)°

m(ZR) = 145°

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The total width for the figure is given as follows:

13 ft 2 in.

The total width for the figure is obtained applying the proportions in the context of the problem.

The measures are given as follows:

The sum of the measures is given as follows:

Each feet is composed by 12 inches, hence:

26 inches = 2 feet and 2 inches.

Hence the sum is given as follows:

11 + 2 = 13 feet and 2 inches.

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Each pair of angles has been identified as adjacent, complementary, supplementary, or vertical angles as shown in the image attached below.

In Mathematics and Geometry, a complementary angle can be defined as two (2) angles or arc whose sum is equal to 90 degrees (90°);

55 + 35 = 90°

In Mathematics and Geometry, adjacent angles can be defined as two (2) angles that share a common vertex and a common side. This ultimately implies that, both angles 1 and 2, 5 and 6, 9 and 10 are pair of adjacent angles.

In conclusion, the linear pair theorem is sometimes referred to as linear pair postulate (supplementary angle) and it states that the measure of two angles would add up to 180° provided that they both form a linear pair.

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The 7.34 cubic yards of concrete needed to pour the foundation walls, which is calculate the volume of the walls and then add the waste factor of 6%.

To find the volume of the concrete needed to pour the foundation walls, we first need to find the total area of the walls. We can do this by breaking it down into three sections

The two 26 ft x 1 ft walls

Area = 2 x (26 ft x 1 ft) = 52 sq ft

The two 42 ft x 1 ft walls

Area = 2 x (42 ft x 1 ft) = 84 sq ft

The 12 in x 15 in x 12 ft wall

First, we need to convert the dimensions to feet:

Length = 12 in ÷ 12 = 1 ft

Breadth = 15 in ÷ 12 = 1.25 ft

Height = 12 ft

Area = (1 ft + 1.25 ft) x 2 x 12 ft = 51 ft²

Total area of the walls = 52 sq ft + 84 sq ft + 51 sq ft = 187 sq ft

Now, we need to add the waste factor of 6% to this to account for any material that may be lost or wasted during the pouring process

Total area with waste factor = 187 sq ft + 6% of 187 sq ft = 198.22 sq ft

Finally, we need to calculate the volume of concrete needed, assuming a thickness of 1 ft

Volume = area x thickness = 198.22 sq ft x 1 ft = 198.22 cubic ft

Since 1 cubic yard is equal to 27 cubic feet, we can convert the volume to cubic yards

Volume in cubic yards = 198.22 cubic ft ÷ 27 = 7.34 cubic yards

Therefore, we need 7.34 cubic yards of concrete to pour the foundation walls with a 6% waste factor.

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Answer:

The asymptote is 0.

Step-by-step explanation:

In [tex]f(x)=a^x+b[/tex], b is the asymptote.

Answer:

it's 0

Step-by-step explanation:

The constant of proportionality for the graph given can be found to be 2 / 3.

A fixed numerical quantity that links two variables exhibiting direct proportionality is referred to as the constant of proportionality. This implies that when two factors are directly proportional, a stable ratio exists between them. The same value defines this figure and is identified as the constant of proportionality.

Pick a point on the graph such as ( 3 , 2 ) and ( 6, 4 ), the constant of proportionality would be:

= Change in y / Change in x

= ( 4 - 2) / ( 6 - 3 )

= 2 / 3

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The simplest form of the expression 6x - 4y - 2x / 2 is 5x - 4y.

What is Simplest form?

In mathematics, simplest form refers to the expression that has been simplified or reduced as much as possible. This means that no further simplification or reduction can be done without changing the value of the expression.

The expression 6x - 4y - 2x / 2 can be simplified using the order of operations (PEMDAS) as follows:

6x - 4y - 2x / 2

= 6x - 4y - x (since 2x / 2 = x)

= 5x - 4y

Therefore, the simplest form of the expression 6x - 4y - 2x / 2 is 5x - 4y.

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Answer:  To show that the equation "3sin () tan () = 5cos () - 2" is equivalent to the equation "(4 cos() - 3)(2 cos () + 1) = 0", we need to simplify the first equation and check if it has the same solutions as the second equation.

Starting with the first equation:

3sin () tan () = 5cos () - 2

Using the identity tan () = sin () / cos (), we can write:

3sin () (sin () / cos ()) = 5cos () - 2

Multiplying both sides by cos (), we get:

3sin^2 () = (5cos () - 2)cos ()

Using the identity sin^2 () + cos^2 () = 1 and rearranging, we get:

3(1 - cos^2 ()) = 5cos^2 () - 2cos ()

Expanding and rearranging, we get:

5cos^2 () - 2cos () - 3 + 3cos^2 () = 0

Simplifying, we get:

8cos^2 () - 2cos () - 3 = 0

Now, we can use the quadratic formula to solve for cos ():

cos () = [2 ± sqrt(2^2 - 4(8)(-3))]/(2(8))

cos () = [2 ± sqrt(100)]/16

cos () = (1/4) or (-3/8)

Substituting these values back into the original equation, we can verify that they satisfy the equation.

Now, let's consider the second equation:

(4 cos() - 3)(2 cos () + 1) = 0

This equation is satisfied when either 4cos() - 3 = 0 or 2cos() + 1 = 0.

Solving for cos() in the first equation, we get:

4cos() - 3 = 0

cos() = 3/4

Substituting this value back into the original equation, we can verify that it satisfies the equation.

Solving for cos() in the second equation, we get:

2cos() + 1 = 0

cos() = -1/2

Substituting this value back into the original equation, we can also verify that it satisfies the equation.

Therefore, we have shown that the equation "3sin () tan () = 5cos () - 2" is equivalent to the equation "(4 cos() - 3)(2 cos () + 1) = 0".

The height of the pole is 18.18 feet

Given data ,

Elka uses similar triangles to find the height of the pole. The situation forms two similar right triangles: one with the pole and the mirror, and another with Elka's eyes and the mirror.

Let's denote the height of the pole as "h" feet.

Distance from the base of the pole to the mirror = 20 feet

Distance Elka moves back away from the mirror = 5.5 feet

Height of Elka's eyes above the ground = 5 feet

Using the concept of similar triangles, we can set up the following proportion:

h / 20 = 5 / 5.5

Multiply by 20 on both sides , we get

h = 100 / 5.5

h = 18.18 feet

Hence , the height of the flagpole is approximately 18.18 feet

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The point P ( 1 , -12 ) lies on the graph of the function f ( x ) = 3x² - 6x - 9

Given data ,

Let the function be represented as f ( x )

Now ,

Let the point be P ( 1 , -12 )

And , to determine if the point (1, -12) is on the graph of the function f(x) = 3x² - 6x - 9, we can substitute x = 1 and y = -12 into the equation and check if it satisfies the equation.

Plugging in x = 1 into the equation, we get:

f(1) = 3(1)² - 6(1) - 9

f(1) = 3 - 6 - 9

f(1) = -12

Hence , when x = 1, f(x) = -12. Since f(1) = -12 and the given point is (1, -12), the point (1, -12) does lie on the graph of the function f(x) = 3x² - 6x - 9

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An equation that shifts the graph of the function f(x) = √x to the left 5 units is: f(x) = √(x + 5)

There are different methods of transformation of functions or graphs and they are:

1) Translation

2) Reflection

3) Dilation

4) Rotation

Now, we can shift a function upwards, downwards, to the left or right as the case may be.

In this case we want to shift the function to the left by 5 units.

Shifting the function 5 units to the left means translating the function 5 units along the x-axis. So we will add 5 to x to get:

f(x) = √(x + 5)

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The difference between the actual mean time taken to complete the quiz and the time Sam says is -0.2 minutes.

The mean, also known as the average, is a measure of central tendency that represents the sum of all values in a data set divided by the number of data points.

Mathematically, the mean is defined as:

Mean = (Sum of all values) / (Number of data points)

According to the given information:

To find the difference between the actual mean time taken to complete the quiz and the time Sam says, we can follow these steps:

Find the actual mean time taken to complete the quiz:

Add up all the given times: 13 + 16 + 8 + 9 + 12 + 14 + 20 + 36 + 11 + 5 + 13 + 16 + 8 + 9 + 12 + 14 + 20 + 36 + 11 + 5 = 236

Divide the sum by the number of data points (20, since there are 20 quiz times): 236 / 20 = 11.8 minutes

Calculate the difference between the actual mean time and the time Sam says:

Actual mean time taken to complete the quiz: 11.8 minutes

Time Sam says: 12 minutes

Difference = Actual mean time - Time Sam says = 11.8 - 12 = -0.2 minutes

So, the difference between the actual mean time taken to complete the quiz and the time Sam says is -0.2 minutes, indicating that Sam's estimate is 0.2 minutes (or 12 seconds) higher than the actual mean time.

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If will be positive if:

-24(a-t) > 0

-(a-t) > 0

-a + t > 0

t > a

Using derivatives, it is found that the particle is moving to the right for t > a , that is, values of t in the interval (a, ∞)

A particle is moving to the right if its velocity is positive.

The position of the particle is given by:

x(t) = 12(a -t)^2

The velocity is the derivative of the position, hence:

v(t) = -24(a-t)

If will be positive if:

-24(a-t) > 0

-(a-t) > 0

-a + t > 0

t > a

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The equivalent speed in still water to achieve the actual speed is 27.172 km/h

Given data ,

The boat's velocity vector in still water will have two components: one along the north direction (opposite to the current) with magnitude "v" km/h, and one along the northwest direction (due to the current) with magnitude 4 km/h.

Since the boat is moving in a direction that is 45 degrees between north and northwest, we can use trigonometry to find the component of velocity in the northwest direction. By using the cosine of 45 degrees, we get:

Component of velocity in northwest direction = 4 km/h x cos(45 degrees) = 4 km/h x 0.707 = 2.828 km/h

Now , adding the components of velocity in the north and northwest directions should result in a speed of 30 km/h, which is the real speed needed to go through still water.

v + 2.828 = 30

v = 27.172 km/h

Hence , the equivalent speed of the boat in still water to achieve an actual speed of 30 km/h is approximately 27.172 km/h

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The solutions to the equation (x - 5)² = 25 are x = 10 and x = 0.

An equation is a mathematical statement that shows that two expressions are equal. It consists of two sides separated by an equal sign (=). Each side of the equation can contain variables, constants, and mathematical operations such as addition, subtraction, multiplication, and division.

It is a shape with four sides of equal length and four right angles. It is also a number multiplied by itself.

Solve the equation (x - 5)² = 25:

1. Take the square root of both sides of the equation to remove the exponent of 2 on the left side.

  √[(x - 5)²] = √25

2. Simplify the left side by removing the exponent of 2 and keeping the absolute value.

  |x - 5| = 5

3. Write two separate equations to account for both possible values of x when taking the absolute value.

  x - 5 = 5   or   x - 5 = -5

4. Solve for x in each equation.

x = 10   or   x = 0

  So the solutions to the equation (x - 5)² = 25 are x = 10 and x = 0.

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Answer:

7.95

Step-by-step explanation:

We can work percentage problems using the formula

P%x = y, where P is the percentage, x is the "of" value in the problem, and y is the "is" value in the problem.

Thus, in the formula, our p value is 0.30, our x ("of") value is 26.5 and we're trying to find our y ("is") value:

0.30 * 26.5 = y

7.95 = y

Therefore, 30% of 26.5 is 7.95

Answer:

I worked this out & I got a horribly messy number, but if you still want it, here you go.

The answer I got is 49.833333333333333333333333333333, or
49 833333333333333333333333333333/100000000000000000000000000000.

I could not simplify it. Hopefully, your teacher accepts this.

let's firstly convert the mixed fractions to improper fractions, then multiply.

[tex]\stackrel{mixed}{8\frac{2}{3}}\implies \cfrac{8\cdot 3+2}{3}\implies \stackrel{improper}{\cfrac{26}{3}}~\hfill \stackrel{mixed}{5\frac{3}{4}} \implies \cfrac{5\cdot 4+3}{4} \implies \stackrel{improper}{\cfrac{23}{4}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{26}{3}\cdot \cfrac{23}{4}\implies \cfrac{26}{4}\cdot \cfrac{23}{3}\implies \cfrac{13}{2}\cdot \cfrac{23}{3}\implies \cfrac{299}{6}\implies 49\frac{5}{6}~ft^2[/tex]

Answer:

choice 1

Step-by-step explanation:

62 and x are a linear pair and sum to 180 , that is

62 + x = 180 ( subtract 62 from both sides )

x = 118

Answer:

1. 118

Step-by-step explanation:

Given the image provided:

Solve for x:

Since, the image is of a supplementary angle that adds up to 180° and one angle is 62°, then we take 180 minus 62.

Answer:

Therefore, x = 118 and the answer is 1.

a. Weight of 25 gallons of water [tex]= (25 \times 25)/(3) = 625/3 lbs[/tex]  B. Weight of the empty tank  [tex]= (625 \times 1)/(3 \times 10) = 625/30 lbs[/tex] C. Weight of tank filled with water[tex]= 229 5/6 lbs[/tex]

a. The weight of 25 gallons of water can be calculated by multiplying the weight of one gallon of water by 25.

Given:

Weight of 1 gallon of water = 8 1/3 lbs

Calculation:

Weight of 25 gallons of water = (Weight of 1 gallon of water) × 25

[tex]= 8 1/3 lbs \times 25[/tex]

To perform the multiplication, let's convert the mixed number 8 1/3 to an improper fraction:

[tex]8 1/3 = (8 \times 3 + 1)/3 = 25/3[/tex]

Substituting back into the equation:

Weight of 25 gallons of water [tex]= 25/3 \times 25[/tex]

Now, we can multiply the fractions:

Weight of 25 gallons of water [tex]= (25 \times 25)/(3)= 625/3 lbs[/tex]

b. The weight of the empty 25-gallon tank can be calculated as 1/10 of the weight of water it can hold.

Given:

Weight of 25 gallons of water = 625/3 lbs

Calculation:

Weight of the empty tank = (Weight of 25 gallons of water) × (1/10)

[tex]= (625/3 lbs) \times (1/10)[/tex]

We can simplify the fractions:

Weight of the empty tank [tex]= (625 \times 1)/(3 \times 10)[/tex]

[tex]= 625/30 lbs[/tex]

c. The weight of a 25-gallon tank filled with water would be the sum of the weight of the empty tank (calculated in part b) and the weight of 25 gallons of water (calculated in part a).

Given:

Weight of 25 gallons of water = 625/3 lbs

Weight of empty tank = 625/30 lbs

Calculation:

Weight of tank filled with water = Weight of 25 gallons of water + Weight of empty tank

[tex]= 625/3 lbs + 625/30 lbs[/tex]

We can find a common denominator for the fractions and add them:

Weight of tank filled with water [tex]r = (625 \times 10)/(3 \tiimes 10) + (625)/(30)[/tex]

[tex]= (6250/30) + (625/30)= 6875/30 lbs[/tex]

We can simplify the fraction:

Weight of tank filled with water [tex]= 229 5/6 lbs[/tex]

Therefore,a.Weight of 25 gallons of water [tex]= (25 \times 25)/(3) = 625/3 lbs[/tex]  B. Weight of the empty tank  [tex]= (625 \times 1)/(3 \times 10) = 625/30 lbs[/tex] C. Weight of tank filled with water[tex]= 229 5/6 lbs[/tex]

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a. Weight of 25 gallons of water [tex]= (25 \times 25)/(3) = 625/3 lbs[/tex]  B. Weight of the empty tank  [tex]= (625 \times 1)/(3 \times 10) = 625/30 lbs[/tex] C. Weight of tank filled with water[tex]= 229 5/6 lbs[/tex]

a. The weight of 25 gallons of water can be calculated by multiplying the weight of one gallon of water by 25.

Given:

Weight of 1 gallon of water = 8 1/3 lbs

Calculation:

Weight of 25 gallons of water = (Weight of 1 gallon of water) × 25

[tex]= 8 1/3 lbs \times 25[/tex]

To perform the multiplication, let's convert the mixed number 8 1/3 to an improper fraction:

[tex]8 1/3 = (8 \times 3 + 1)/3 = 25/3[/tex]

Substituting back into the equation:

Weight of 25 gallons of water [tex]= 25/3 \times 25[/tex]

Now, we can multiply the fractions:

Weight of 25 gallons of water [tex]= (25 \times 25)/(3)= 625/3 lbs[/tex]

b. The weight of the empty 25-gallon tank can be calculated as 1/10 of the weight of water it can hold.

Given:

Weight of 25 gallons of water = 625/3 lbs

Calculation:

Weight of the empty tank = (Weight of 25 gallons of water) × (1/10)

[tex]= (625/3 lbs) \times (1/10)[/tex]

We can simplify the fractions:

Weight of the empty tank [tex]= (625 \times 1)/(3 \times 10)[/tex]

[tex]= 625/30 lbs[/tex]

c. The weight of a 25-gallon tank filled with water would be the sum of the weight of the empty tank (calculated in part b) and the weight of 25 gallons of water (calculated in part a).

Given:

Weight of 25 gallons of water = 625/3 lbs

Weight of empty tank = 625/30 lbs

Calculation:

Weight of tank filled with water = Weight of 25 gallons of water + Weight of empty tank

[tex]= 625/3 lbs + 625/30 lbs[/tex]

We can find a common denominator for the fractions and add them:

Weight of tank filled with water [tex]r = (625 \times 10)/(3 \tiimes 10) + (625)/(30)[/tex]

[tex]= (6250/30) + (625/30)= 6875/30 lbs[/tex]

We can simplify the fraction:

Weight of tank filled with water [tex]= 229 5/6 lbs[/tex]

Therefore,a.Weight of 25 gallons of water [tex]= (25 \times 25)/(3) = 625/3 lbs[/tex]  B. Weight of the empty tank  [tex]= (625 \times 1)/(3 \times 10) = 625/30 lbs[/tex] C. Weight of tank filled with water[tex]= 229 5/6 lbs[/tex]

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Answer:

Step-by-step explanation:

Answer:

It will take approximately 6.89 years for the investment to grow to $9737.

Step-by-step explanation:

We can use the formula for compound interest to solve this problem:

A = P(1 + r/n)^(nt)

Where:

A = the future value of the investment

P = the present value of the investment

r = the annual interest rate (as a decimal)

n = the number of times the interest is compounded per year

t = the number of years

We know that P = $7000, r = 0.034, n = 4 (since the interest is compounded quarterly), and we want to find t when A = $9737.

$9737 = $7000(1 + 0.034/4)^(4t)

Divide both sides by $7000:

1.391 = (1 + 0.034/4)^(4t)

Take the natural logarithm of both sides:

ln(1.391) = ln[(1 + 0.034/4)^(4t)]

Use the power rule of logarithms:

ln(1.391) = 4t * ln(1 + 0.034/4)

Divide both sides by 4 ln(1 + 0.034/4):

t = ln(1.391) / [4 * ln(1 + 0.034/4)]

Using a calculator, we find:

t ≈ 6.89 years

Therefore, it will take approximately 6.89 years for the investment to grow to $9737.

The value of evaluating the integral expression [tex]\int\limits^a_{-a} \int\limits^{\sqrt{a^2 - y^2}}_{-\sqrt{a^2 - y^2}} \int\limits^{\sqrt{a^2 -x^2 - y^2}}_{-\sqrt{a^2 - x^2 - y^2}} (x^2z + y^2z + z^3) dz dx dy[/tex] is 0

Given that

[tex]\int\limits^a_{-a} \int\limits^{\sqrt{a^2 - y^2}}_{-\sqrt{a^2 - y^2}} \int\limits^{\sqrt{a^2 -x^2 - y^2}}_{-\sqrt{a^2 - x^2 - y^2}} (x^2z + y^2z + z^3) dz dx dy[/tex]

To change to spherical coordinates, we need to express x, y, and z in terms of spherical coordinates: r, θ, and Φ .

In particular, we have

[tex]x &= r \sin\phi \cos\theta, \\y &= r \sin\phi \sin\theta, \\z &= r \cos\phi[/tex]

The Jacobian for the transformation is r² sin(Φ), and the limits of integration become

[tex]-a &\leq x \leq a \quad \Rightarrow \quad 0 \leq r \leq a, \\-\sqrt{a^2 - y^2} &\leq y \leq \sqrt{a^2 - y^2} \quad \Rightarrow \quad 0 \leq \phi \leq \frac{\pi}{2}, \\-\sqrt{a^2 - x^2 - y^2} &\leq z \leq \sqrt{a^2 - x^2 - y^2} \quad \Rightarrow \quad 0 \leq \theta \leq 2\pi.[/tex]

Substituting into the integral, we have

[tex]&\int_{0}^{a} \int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} (r^2\sin^2\phi\cos\theta\cdot r\cos\phi + r^2\sin^2\phi\sin\theta\cdot r\cos\phi + r^3\cos^3\phi) r^2 \sin\phi,d\theta d\phi dr \[/tex]

[tex]&\quad = \int_{0}^{a} \int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} (r^3\sin^2\phi\cos\theta\cos\phi + r^3\sin^2\phi\sin\theta\cos\phi + r^3\cos^3\phi) \sin\phi, d\theta d\phi dr[/tex]

[tex]&\quad = \int_{0}^{a} \int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} r^3\sin\phi\cos\phi (\sin^2\phi\cos\theta + \sin^2\phi\sin\theta + \cos^2\phi) , d\theta d\phi dr \[/tex]

[tex]&\quad = \int_{0}^{a} \int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} r^3\sin\phi\cos\phi (\sin^2\phi + \cos^2\phi) , d\theta d\phi dr \[/tex]

[tex]&\quad = \int_{0}^{a} \int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} r^3\sin\phi\cos\phi, d\theta d\phi dr \[/tex]

[tex]&\quad = \int_{0}^{a} \int_{0}^{\frac{\pi}{2}} 0, d\theta d\phi dr \&\quad = 0[/tex]

Therefore, the value of the integral is 0.

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The speed s, in feet per second, of the car before the brakes were applied is,

⇒ s = 67.5 m/s

To multiply means to add a number to itself a particular number of times. Multiplication can be viewed as a process of repeated addition.

Given that;

Skid mark for the car = 150 ft

And, given equation of speed and distance is,

⇒ s = √24d

Where, d is the distance

And, s is the speed  

Hence, We get;

s = √ 24 × 150

s = √3600

s = 60 m/s

Thus, The speed of the car before it stop is equal to s = 60 m/s

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Answer: 15g-25h

Step-by-step explanation:

To use distributive property, you multiply 5x3g individually, and then 5x-5h individually.

You then add them.

So:

5(3g-5h)

=15h-25j

Answer:

75 - 25π.

Step-by-step explanation:

To find the work done by a force field on a particle moving along a curve, we use the line integral of the force field over that curve.

In this case, the curve is a circle of radius 5 centered at (0, 0, 3) lying on the plane z = 3. We can parameterize this curve using polar coordinates as:

r(t) = (5cos(t), 5sin(t), 3), where t goes from 0 to 2π.

The differential of the curve, dr(t), is given by:

dr(t) = (-5sin(t), 5cos(t), 0) dt

Now we need to calculate the work done by the force field F along this curve. The line integral of F over the curve is given by:

W = ∫ F · dr = ∫ (2x +y²Z)dx + (2x-4y+Z)dy + (x-2y-Z²)dz

Substituting x = 5cos(t), y = 5sin(t), and z = 3, we get:

W = ∫ (10cos(t) + 25sin²(t)·3) (-5sin(t))dt

∫ (10cos(t) - 20sin(t) + 3) (5cos(t))dt

∫ (5cos(t) - 10sin(t) - 9) (0)dt

Simplifying, we get:

W = -75∫sin(t)cos(t)dt + 50∫cos²(t)dt + 0

Using the trigonometric identities sin(2t) = 2sin(t)cos(t) and cos²(t) = (1 + cos(2t))/2, we can simplify this further:

W = -75∫(1/2)sin(2t)dt + 25∫(1 + cos(2t))dt

= -75·(1/2)·(-cos(2t))∣₀^(2π) + 25·(t + (1/2)sin(2t))∣₀^(2π)

= 75 - 25π

Therefore, the work done by the force field F on the particle moving along the circle of radius 5 centered at (0, 0, 3) lying on the plane z = 3 is 75 - 25π.

The sample of given data is Sample 1: 1, 2, 3, 4, 5 Sample 2: 6, 7, 8, 9, 10

b)Sample 1: 1, 2, 3, 4, 5 Sample 2: 6, 7, 8, 9, 10

(a) An example of data with SS(between) = 0 and SS(within) > 0 could be the following:

Sample 1: 1, 2, 3, 4, 5

Sample 2: 6, 7, 8, 9, 10

Sample 3: 11, 12, 13, 14, 15

In this example, the means of each sample are all different from each other, but the grand mean (8) is equal to the mean of each sample. Therefore, there is no variability between the means of the samples, resulting in SS(between) = 0. However, there is still variability within each sample, resulting in SS(within) > 0.

(b) An example of data with SS(between) > 0 and SS(within) = 0 could be the following:

Sample 1: 1, 2, 3, 4, 5

Sample 2: 6, 7, 8, 9, 10

Sample 3: 11, 12, 13, 14, 15

In this example, the means of each sample are all the same (8), but the values within each sample are all different from each other. Therefore, there is variability between the means of the samples, resulting in SS(between) > 0. However, there is no variability within each sample, resulting in SS(within) = 0.

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