Solve cos x +sqr root of 2 = -cos x for x over the interval 0,2pi

To find the slope of a line perpendicular to the line defined by the equation 0.5x−5y=9, first convert the equation into slope-intercept form to determine the slope of the original line. The slope of the line perpendicular to this is its negative reciprocal. For this specific problem, the slope of the line perpendicular to the given line is -10.

Firstly, rearrange the given equation, 0.5x−5y=9, into the slope-intercept format, i.e. y=mx+b. Here, 'm' is the slope of the line. To illustrate this rearrangement: 5y = 0.5x - 9, then divide through by 5 to get y = 0.1x - 1.8. Thus, the slope of the given line is 0.1.

Lines that are perpendicular to each other have slopes that are negative reciprocals of each other. In this case, the slope of the given line is 0.1, so the negative reciprocal (which represents the slope of the line perpendicular to this line) is -1/0.1, which equals -10. This is the slope of the line perpendicular to the given line.

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Answer:

Option B.

Step-by-step explanation:

We need [tex]2\frac{3}{4}[/tex] wheelbarrows of sand for 8 wheelbarrows of concrete.

That means ratio between sand and concrete is [tex]2\frac{3}{4}:8[/tex]

Or [tex]\frac{11}{4}:8[/tex]

Now we have to calculate the amount of sand to make 248 cubic feet of concrete.

If the amount of sand required is x cubic feet then the ratio of sand and concrete will be x : 248.

Since both the ratios should be same therefore,

[tex]\frac{x}{248}=\frac{\frac{11}{4} }{8}[/tex]

[tex]\frac{x}{248}=\frac{11}{32}[/tex]

x = [tex]\frac{11\times 248}{32}[/tex]

x = [tex]\frac{2728}{32}[/tex]

x = [tex]\frac{341}{4}[/tex]

  = [tex]85\frac{1}{4}[/tex] cubic feet

Option B will be the answer.

Answer:

$735

Step-by-step explanation:

$180/week * 2 weeks = $360 for 2 weeks

$0.75/mile * 500 miles = $375

$360 + $375 = $735 for 2 week trip of 50 miles

To solve this problem, let us first calculate for the Perimeter of the other octagon. The formula for Perimeter is:

Perimeter = n * l

Where n is the number of sides (8) and l is the length of one side. Let us say that first octagon is 1 and the second octagon is 2 so that:

Perimeter 2 = 8 * 16.35 in = 130.8 inch

We know that Area is directly proportional to the square of Perimeter for a regular polygon:

Area = k * Perimeter^2

Where k is the constant of proportionality. Therefore we can equate 1 and 2 since k is constant:

Area 1 / Perimeter 1^2 = Area 2 / Perimeter 2^2

Substituting the known values:

392.4 inches^2 / (87.2 inch)^2 = Area 2 / (130.8 inch)^2

Area 2 = 882.9 inches^2

Therefore the area of the larger octagon is about 882.9 square inches.

The width of the photograph is 8 inches

let

width of the photograph = x

length of the photograph = 6 + x

Area  = x(6 + x)

112 =  x(6 + x)

112 = 6x + x²

x² + 6x - 112 = 0

x² - 8x + 14x - 112 = 0

x(x - 8) + 14(x - 8) = 0

(x + 14)(x - 8)

x = -14 or 8

x = 8

x cannot be negative so we use only 8.

width of the photograph = 8 inches

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The provided table outlines the probability distribution for drawing varying numbers of aces in a random draw of 4 cards from a standard 52-card deck, using combinatorial probability calculations.

To answer your question, we need to consider the different possibilities for pulling aces in a draw of four cards. There are 4 aces in a standard deck of 52 cards, and here's a table showing the probability distribution for each outcome:

Please note that 'comb(a, b)' in this case represents 'a choose b', which is a way to compute combinations in probability.

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To find the volume of the solid that lies between the paraboloid z = x^2 + y^2 and the sphere x^2 + y^2 + z^2 = 2 using cylindrical coordinates, set up the integral for the volume by rewriting the sphere equation in cylindrical coordinates, determining the limits for r and z, and evaluating the integral.

To find the volume of the solid that lies between the paraboloid z = x^2 + y^2 and the sphere x^2 + y^2 + z^2 = 2

We can rewrite the sphere equation in cylindrical coordinates as r^2 + z^2 = 2. The limits for r are from 0 to √(2-z^2), and for z, they are from 0 to √(2-r^2).

The volume can be found by integrating the constant 1 over the limits of r and z: V = ∭1 dz dr dθ. Evaluate this integral to find the volume of the solid.

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The least common multiple of 2, 10 and 6 by using the definition of multiple is 30.

The lowest possible number that  can be divisible by all the given numbers is called as  least common multiple (LCM). It is the smallest multiple which is common in all the numbers.

The least common multiple of 2, 10 and 6 can be calculated as:

The multiples of 2 are: 2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34.........

The multiples of 10 are: 10,20,30,40,50,60,70.............

The multiples of 6 are: 6,12,18,24,39,36,42,48,54,60,66...........

The lowest common multiple among 2,10,6 is 30.

Thus, the least common multiple of 2, 10 and 6 is 30.

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there are 31 days in August

 365 days in a year

31/365 = 0.0849 probability

Answer:

0.0849

Step-by-step explanation:

Let's consider the event: a birthday takes place in March. The probability (P) of such event is:

[tex]P = \frac{favorable\ cases }{possible\ cases}[/tex]

The favorable cases are 31 because March has 31 days.

The possible cases are 365 because there are 365 days in a year.

The probability of a birthday being in March is:

[tex]P=\frac{31}{365} =0.0849[/tex]

Step 1: The intersection curve is only at the origin.

Step 2: Set up the volume integral over the sphere region.

Step 3:Evaluate the integral to find the volume: [tex]\(V = 216\pi\sqrt{2}\).[/tex]

To find the volume of the solid enclosed by the cone [tex]\(z = x^2 + y^2\)[/tex] and the sphere[tex]\(x^2 + y^2 + z^2 = 72\),[/tex] we'll first find the intersection curve of the cone and the sphere in cylindrical coordinates, then set up the triple integral to find the volume.

Step 1: Finding the intersection curve:

The cone equation in cylindrical coordinates becomes [tex]\(z = r^2\)[/tex] and the sphere equation remains the same as [tex]\(r^2 + z^2 = 72\).[/tex]

To find the intersection, we set these two equations equal to each other:

[tex]\[r^2 = r^2 + z^2\][/tex]

Substitute [tex]\(z = r^2\)[/tex] from the cone equation:

[tex]\[r^2 = r^2 + (r^2)^2\][/tex]

[tex]\[r^2 = r^2 + r^4\][/tex]

[tex]\[r^4 = 0\][/tex]

From this, we see that the only solution is (r = 0), which corresponds to the point at the origin.

Step 2: Setting up the integral for volume:

We'll integrate over the region where the cone lies within the sphere, which is the entire sphere. In cylindrical coordinates, the limits of integration are [tex]\(0 \leq r \leq 6\)[/tex] (since the sphere has radius [tex]\(\sqrt{72} = 6\)) and \(0 \leq \theta \leq 2\pi\).[/tex]

The limits for \(z\) are from the cone to the sphere, so it's from [tex]\(r^2\) to \(\sqrt{72-r^2}\).[/tex]

Thus, the volume integral is:

[tex]\[V = \iiint_{E} r \, dz \, dr \, d\theta\][/tex]

Where (E) is the region enclosed by the cone and the sphere.

Step 3: Evaluate the integral:

[tex]\[V = \int_{0}^{2\pi} \int_{0}^{6} \int_{r^2}^{\sqrt{72-r^2}} r \, dz \, dr \, d\theta\][/tex]

Let's evaluate this integral step by step:

[tex]\[V = \int_{0}^{2\pi} \int_{0}^{6} (r\sqrt{72-r^2} - r^3) \, dr \, d\theta\][/tex]

[tex]\[V = \int_{0}^{2\pi} \left[-\frac{1}{4}(72-r^2)^{3/2} - \frac{1}{4}r^4\right]_{0}^{6} \, d\theta\][/tex]

[tex]\[V = \int_{0}^{2\pi} \left[-\frac{1}{4}(0)^{3/2} - \frac{1}{4}(6^4) - \left(-\frac{1}{4}(72)^{3/2} - \frac{1}{4}(0)^4\right)\right] \, d\theta\][/tex]

[tex]\[V = \int_{0}^{2\pi} \left[\frac{1}{4}(72)^{3/2}\right] \, d\theta\][/tex]

[tex]\[V = \frac{1}{4}(72)^{3/2} \int_{0}^{2\pi} 1 \, d\theta\][/tex]

[tex]\[V = \frac{1}{4}(72)^{3/2} \cdot 2\pi\][/tex]

[tex]\[V = 36\pi \sqrt{72}\][/tex]

[tex]\[V = 36\pi \times 6\sqrt{2}\][/tex]

[tex]\[V = 216\pi\sqrt{2}\][/tex]

So, the volume of the solid enclosed by the cone and the sphere is [tex]\(216\pi\sqrt{2}\).[/tex]

The area of the larger square is 4 times larger than the area of the smaller square. The area of the big hexagon is 12 sq. in.

The area of the larger square is 4 times larger than the area of the smaller square. This is because the area of a square is proportional to the square of its side length.

In this case, the side length of the larger square is twice the side length of the smaller square, so the area of the larger square is 2² times greater than the area of the smaller square.

Given that the area of the small hexagon is 3 sq. in, the area of the big hexagon can be found by multiplying the area of the small hexagon by the square of the scale factor:

Area of big hexagon = (scale factor)² * Area of small hexagon = 2² * 3 sq. in = 12 sq. in

The solution is $ 153

The change in the net working capital is $ 153

What is Net Working Capital?

The difference between a company's current assets and current or short-term liabilities is known as net working capital, or working capital.

Cash flow will have an operational origin, when there is a net decrease in working capital

Working Capital = Current Assets - Current Liabilities

Given data ,

Let the change in net working capital be A

Now , the equation will be

Working Capital at the beginning = Current Assets - Current Liabilities

Substituting the values in the equation , we get

Working Capital at the beginning = 316 - 220

Working Capital at the beginning = $ 96

And ,

Working Capital at the end = Current Assets - Current Liabilities

Substituting the values in the equation , we get

Working Capital at the end = 469 - 260

Working Capital at the end = $ 209

So ,

The change in net working capital A = Working Capital at the end - Working Capital at the beginning

Substituting the values in the equation , we get

The change in net working capital A = 209 - 96

The change in net working capital A = $ 153

Therefore , the value of A is $ 153

Hence , change in the net working capital is $ 153

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Answer:

10 times greater

Step-by-step explanation:

700÷70=10

Answer:

[tex]150.72 feet^2[/tex] is the difference in the area covered.

Step-by-step explanation:

A lawn sprinkler sprays water 8 feet at full pressure, P.

A lawn sprinkler rotates 360 degree which means area covered by sprinkler is of circular shape. Since the sprinkler is in center and sprays the the water 8 feet away in all the direction while rotating.

Radius of the circle = 8 feet

Maximum pressure = P

As we know that higher the pressure higher will the force by which water will move out of the sprinkler. And with more force, sprinkler will able to spray water farther.

So we this we can say that pressure of the sprinkler is directly proportional to the radius of the circle in which water sprayed

[tex]pressure\propto Radius[/tex]

[tex]P\propto r[/tex]

[tex]\frac{P_1}{r_1}=\frac{P_2}{r_2}=constant[/tex]

[tex]P_1=P.P_2=P-50\%\times P=0.5 P[/tex]

[tex]r_1=8 feet.r_2=?[/tex]

[tex]r_2=\frac{0.5 P\times 8 feet}{P}=4 feet[/tex]

Area when , [tex]r_1= 8 feet[/tex] (Area of circle=[tex]\pi (radius)^2[/tex])

[tex]A=\pi r_1^{2}=\pi (8 feet)^2[/tex]

Area when ,[tex]r_2= 4 feet[/tex]

[tex]A'=\pi r_1^{2}=\pi (4 feet)^2[/tex]

Difference in Area = A- A'

[tex]\pi (8 feet)^2-\pi(4 feet)^2=\pi(48 feet^2)=150.72 feet^2[/tex]

[tex]150.72 feet^2[/tex] is the difference in the area covered.

The angular displacement of the ball while it's in the air is 9.77 radians.

Here's how we can find it:

Time in the air: First, we need to find the time the ball spends in the air. We can use the following kinematic equation for vertical motion:

h = 1/2 * g * t^2

where:

h is the vertical distance (2.08 m)

g is the acceleration due to gravity (9.81 m/s²)

t is the time in the air

Solving for t, we get:

t = sqrt(2 * h / g) = sqrt(2 * 2.08 m / 9.81 m/s²) ≈ 1.43 s

Angular velocity: The angular velocity (ω) of the ball is related to its linear velocity (v) and radius (r) by the following equation:

ω = v / r

where:

v is the linear velocity (3.00 m/s)

r is the radius (0.200 m)

Plugging in the values, we get:

ω = 3.00 m/s / 0.200 m = 15 rad/s

Angular displacement: Finally, the angular displacement (θ) of the ball is the product of its angular velocity (ω) and the time (t) it spends in the air:

θ = ω * t = 15 rad/s * 1.43 s ≈ 21.45 rad ≈ 9.77 rad (rounded to three significant figures)

Therefore, the angular displacement of the ball while in the air is approximately 9.77 radians.

Answer:

inceases by factor four

Step-by-step explanation: