A brick of mass 2 kg falls through water with an acceleration of 2 ms 2. The total force of the resistance is N.​

The vector equations of L₂ when expressed as Cartesian equations are

y = 0

x = (z ± √(z² - 4(z-2))) / 2

z = [(4 - ((x-1)/(-z))²) ± √((4 - ((x-1)/(-z))²)² + 64)] / 8

To find the vector equation of the line L₂, we need to find a vector that is perpendicular to L₁ and passes through the point (1,0,2). Let's start by finding the vector equation of L₁.

Let P(x,y,z) be a point on L₁. Then the vector equation of L₁ is given by:

r₁ = P + t * d₁

where d₁ is the direction vector of L₁ and t is a scalar parameter.

Since L₂ is perpendicular to L₁, its direction vector must be perpendicular to d₁. Thus, we can find a vector that is perpendicular to d₁ by taking the cross product of d₁ with any non-zero vector that is not parallel to d₁. Let's choose the vector (0,1,0):

v = d₁ x (0,1,0) = (-z,0,x)

Note that we can choose any non-zero vector that is not parallel to d₁, and we will still get a vector that is perpendicular to d₁.

Now we have a point on L₂ (1,0,2) and a direction vector (v), so we can write the vector equation of L₂:

r₂ = (1,0,2) + s * v

where s is a scalar parameter.

To express the Cartesian equations of L₂, we can write the vector equation as a set of three parametric equations:

x = 1 - sz

y = 0

z = 2 + sx

We can eliminate the parameter s by solving for it in two of the equations and substituting into the third equation:

s = (x - 1) / (-z)

s = (z - 2) / x

Setting these two expressions equal to each other and solving for x, we get:

[tex]x^2 - zx + z - 2 = 0[/tex]

This is a quadratic equation in x, so we can solve for x using the quadratic formula:

[tex]x = (z \± \sqrt{(z^2 - 4(z-2)})) / 2[/tex]

Substituting this expression for x into one of the parametric equations, we get:

y = 0

And substituting the expressions for x and s into the other parametric equation, we get:

[tex]z = 2 + [(z \± \sqrt{(z^2 - 4(z-2)})) / 2] * [(1 - sz) / (-z)][/tex]

Simplifying this equation, we get:

[tex]4z^2 - (4 - s^2)z - 4 = 0[/tex]

Again, this is a quadratic equation in z, so we can solve for z using the quadratic formula:

[tex]z = [(4 - s^2) \± \sqrt((4 - s^2)^2 + 64)] / 8[/tex]

z = [(4 - s²) ± √((4 - s²)² + 64)] / 8

Finally, we can substitute these expressions for x and z into one of the parametric equations to get:

[tex]y = 0\\x = (z \± \sqrt{(z^2 - 4(z-2)})) / 2\\z = [(4 - ((x-1)/(-z))^2) \± \sqrt{((4 - ((x-1)/(-z))^2)^2} + 64)] / 8[/tex]

These are the Cartesian equations of L₂.

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Answer:

C

Step-by-step explanation:

Variability - Lack of consistency or fixed pattern; liability to vary or change

Every student most likely won’t have the same amount of letters in their first name.

according the given question the exact value of given expression is [tex]$\cos\frac{x}{2} = -\sqrt{\frac{1}{2(1 - \left(-\frac{160}{81}\right)^2)}} = -\sqrt{\frac{81^2}{2(81^2 - 160^2)}} = \boxed{-\frac{81\sqrt{239}}{319}}$[/tex]

First, we need to find [tex]$\sin x$[/tex] using the identity[tex]$\cos^2x + \sin^2x = 1$:$\sin^2x = 1 - \cos^2x = 1 - \left(-\frac{4}{5}\right)^2 = \frac{9}{25}$[/tex]

Since  [tex]$\frac{\pi}{2} < x < \pi$[/tex], we know that  [tex]$\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$[/tex]. Therefore, we can use the

identity [tex]$\tan\frac{x}{2} = \frac{\sin x}{1 + \cos x}$[/tex]:

[tex]$\tan\frac{x}{2} = \frac{\sqrt{\frac{9}{25}}}{1 - \frac{4}{5}} = \frac{\frac{3}{5}}{\frac{1}{5}} = \boxed{3}$[/tex]

[tex]If $\tan x = \frac{40}{9}$ and $\pi < x < \frac{3\pi}{2}$, find $\cos\frac{x}{2}$.[/tex]

First, we need to find [tex]$\sin x$[/tex] using the identity [tex]$\tan^2x + 1 = \sec^2x$[/tex]:

[tex]$\sin x = \frac{\tan x}{\sec x} = \frac{\frac{40}{9}}{-\frac{9}{40}} = -\frac{160}{81}$[/tex]

[tex]Since $\pi < x < \frac{3\pi}{2}$, we know that $\frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$[/tex]. Therefore, we can use the identity [tex]$\cos\frac{x}{2} = \pm\sqrt{\frac{1 + \cos x}{2}}$[/tex]:

[tex]$\cos\frac{x}{2} = -\sqrt{\frac{1 + \cos x}{2}} = -\sqrt{\frac{1 + \frac{\cos^2x}{\sin^2x}}{2}} = -\sqrt{\frac{\sin^2x + \cos^2x}{2\sin^2x}} = -\sqrt{\frac{1}{2(1 - \sin^2x)}}$[/tex]

Plugging in [tex]$\sin x = -\frac{160}{81}$[/tex] , we get:

[tex]$\cos\frac{x}{2} = -\sqrt{\frac{1}{2(1 - \left(-\frac{160}{81}\right)^2)}} = -\sqrt{\frac{81^2}{2(81^2 - 160^2)}} = \boxed{-\frac{81\sqrt{239}}{319}}$[/tex]

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Answer:

Both data sets represent multiplicative relationships. In Table A, y is 3.5 times x, and in Table B, y is 9.2 times x.

The correct statement is: Both data sets represent multiplicative relationships. In Table A, y is 3.5 times x, and in Table B, y is 9.2 times x.

Answer:

450

Step-by-step explanation:

Miles is going to spend less than the hours that were spent last week since 2/3 is a fraction.

A fraction represents a part of a whole, and is therefore always less than the whole. For example, 2/3 represents two out of three equal parts of a whole.

The implication of this is that the time that Miles would have to spend on biking in the coming week would be wo out of three equal parts of a whole time that was spent in the last week and this would be less than the time spent last week.

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The best linear equation that describe given graph is y=2.5x-35 i.e. C.

What is a linear equation ?

A linear equation is a mathematical equation that, when plotted on a graph, represents a straight line. It is an equation of the following form:

y = mx + b

where y = dependent variable, x = independent variable, m= slope of the line, and b = y-intercept (the point where the line crosses the y-axis).

Now,

Lets take a point on the given graph that is on line (50,90)

then the equation of line should give the same values.

So,

For A,  y=0.4x-35

y=0.4*50-35

y=-15  Hence, It does not follow.

For B, y= y=0.4x-70

y=0.4*50-70

y=-50  Hence, It does not follow.

For C, y= y=2.5x-35

y=2.5*50-35

y=125-35

y=90

 Hence, It does follow the given point.

Therefore, The best linear equation that describe given graph is y=2.5x-35

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The three-dimensional object in the figure has a surface area of 52 square yards.

Surface area is the two-dimensional measure or area of a three-dimensional space's boundary, just as perimeter is the one-dimensional measure or length of a two-dimensional area's boundary.

The object's rectangular base has a length of 4 yards and a width of 3 yards, as seen in the figure. As a result, the base's area is:

Area of base = length x width = 4 x 3 = 12 square yards

Area of a triangular face = (1/2) x base x height = (1/2) x 4 x 5 = 10 square yards

Since the object has four triangular faces, the total area of the four triangular faces is:

Total area of triangular faces = 4 x 10 = 40 square yards

We add the base's area to the sum of the four triangle sides to determine the object's surface area:

Surface area = Area of base + Total area of triangular faces

Surface area = 12 + 40 = 52 square yards

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The measures οf the three angIes are x = 51.43 degrees, y = 92.85 degrees, and z = 118.85 degrees.

A Iinear equatiοn is a mathematicaI equatiοn that describes a straight Iine in a twο-dimensiοnaI pIane.

We can use the infοrmatiοn given in the prοbIem tο fοrm a system οf three equatiοns with three variabIes. Let x, y, and z represent the measures οf the first, secοnd, and third angIes, respectiveIy.

Frοm the first piece οf infοrmatiοn, we knοw that: x + y + z = 180

Frοm the secοnd piece οf infοrmatiοn, we knοw that: y + z = 5x

Frοm the third piece οf infοrmatiοn, we knοw that: z = y + 26

We can substitute the third equatiοn intο the secοnd equatiοn tο eIiminate z:

y + (y + 26) = 5x

2y + 26 = 5x

2y = 5x - 26

y = (5x - 26)/2

We can substitute this expressiοn fοr y intο the first equatiοn tο eIiminate y and z:

x + (5x - 26)/2 + (5x - 26)/2 + 26 = 180

2x + 5x - 26 + 26 = 360

7x = 360

x = 51.43

We can substitute this vaIue οf x back intο the expressiοn fοr y tο find y:

y = (5x - 26)/2

y = (5(51.43) - 26)/2

y = 92.85

FinaIIy, we can use the equatiοn z = y + 26 tο find z:

z = y + 26

z = 92.85 + 26

z = 118.85

Therefοre, the measures οf the three angIes are x = 51.43 degrees, y = 92.85 degrees, and z = 118.85 degrees.

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Step-by-step explanation:

remember the trigonometric triangle inscribed in a circle (the norm circle with radius = 1).

for a larger system the leg lengths are sine and cosine multiplied by the radius (which is the Hypotenuse of the right-angled triangle).

the longer leg is opposite of the larger angle (60°).

the shorter leg is opposite of the smaller angle (30°).

what is the desired solution ? the length of the Hypotenuse ? or what ?

6 = sin(60)×Hypotenuse

Hypotenuse = 6/sin(60) = 6/(sqrt(3)/2) = 12/sqrt(3) =

= 6.92820323... ≈ 6.9

The angles and lines that match the angles and lines in the diagram consisting of two lines and their common transversal to the correct description  are;

Alternate Interior Angles; 2. ∠4 and ∠8

Consecutive Exterior  Angles; 5. ∠1 and ∠6

Alternate Exterior Angles; 3. ∠1 and ∠5

Transversal; line l

Consecutive Interior Angles; 4. ∠3 and ∠4

Corresponding Angles; 1. ∠1 and ∠7

A transversal is a line that intersects two or more other lines.

The description of the relationship between the angles in the question are;

Alternate Interior Angles

Alternate interior angles are a pair of angles formed when a transversal intersects two lines. They are located between the two lines on opposite side of the transversal.

Consecutive Exterior Angles

Consecutive Exterior Angles are a pair of angles formed when a transversal intersects two lines. They are located outside the two lines on the same side of the transversal

Alternate Exterior Angles

Alternate exterior angles are a pair of angles formed when a transversal intersects two lines. They are located outside the two lines on the opposite sides of the transversal.

Consecutive Interior Angles

Consecutive Interior Angles, also known as Same-Side Interior Angles are a pair of angles formed when a transversal intersects two parallel lines. They are located between the two parallel lines on the same side of the transversal.

Corresponding Angles

Corresponding angles are a pair of angles formed when a transversal intersects two lines. They are located on the same relative positions with respect to the transversal and the two lines.

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The two statements that are both true are as follows: line

AC is perpendicular to line HB and line AC is parallel to FG. That is option A.

A perpendicular line is defined as the line that forms angle 90° where it meets with another line in a plane.

A line is said to be parallel to each other when they do not intercept as they are both on the same plane.

From the given diagram, line AC is perpendicular to line HB because they form angle 90° at the point of intersection.

Also, line AC is parallel to FG, because they can never intersect till infinity.

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The surface area of an ice cube with 4 cm sides is 96 square centimeters.

It is a measurement of the sum of all the areas of the faces, sides, and any other surfaces of an object. For example, a cube has six square faces, each with the same area.

To find the surface area of an ice cube with 4 cm sides, we need to add up the areas of all six faces. Each face is a square with side length 4 cm, so the area of each face is:

Area of one face = (side length)² = 4 cm × 4 cm = 16 cm²

Since there are six faces on a cube, the total surface area of the ice cube is:

Total surface area = 6 × (area of one face) = 6 × 16 cm² = 96 cm²

Therefore, the surface area of an ice cube with 4 cm sides is 96 square centimeters.

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Answer:

The area remaining, correct to the nearest square meter, is approximately 436 square meters.

Step-by-step explanation:

To find the area remaining, we need to subtract the area of the flower bed and the area of the pool from the total area of the garden.

The total area of the garden is:

40m x 15m = 600 square meters

The flower bed is 1m wide and runs along two shorter sides and one long side of the garden. So the area of the flower bed is:

(40m + 2 x 1m) x (15m + 2 x 1m) - 40m x 15m

= (42m x 17m) - (40m x 15m)

= 714 - 600

= 114 square meters

Now let's calculate the area of the pool. The diameter of the pool is 8m, so the radius is 4m. The area of the pool is:

π x (4m)^2

= 16π

≈ 50.27 square meters (rounded to two decimal places)

So the area remaining is:

600 square meters - 114 square meters - 50.27 square meters

≈ 435.73 square meters

Therefore, the area remaining, correct to the nearest square meter, is approximately 436 square meters.

Area of triangle= 24 cm²

Area of sector= 72.243 cm²

Total area= 96.243 cm².

A polygon with three sides and three vertices is called a triangle. It is one of the fundamental geometric forms. Triangle ABC is the designation for a triangle with points A, B, and C. In Euclidean mathematics, any three points that are not collinear produce a distinct triangle and a distinct plane.

The portion of a disc enclosed by two radii and an arc is called a circular sector, also known as a circle sector or disc sector. The smaller area is referred to as the minor sector, and the bigger area as the major sector. A sector is referred to as a component of a circle made up of the circular's arc and its two radii.

In this question,

Area of triangle= 1/2 × base × height

                         = 1/2 × 6 ×8

                         = 24 cm²

Area of sector = (θ/360°) × πr²

                        = (82/360) × πr²

Here r=√6²+8²

          = 10 cm

Area of sector = (82/360) × π10²

                        = 0.23 × 314.1

                        = 72.243 cm²

Total= 24 cm²+72.243 cm²= 96.243 cm².

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In the polygon, using sum of exterior angles the value of x = 37°

A polygon is a shape that has 3 or more sides.

Given the polygon which is a hexagon to find the value of x, we note that the angles are all exterior angles. We know that the sum of the exterior angles of a polygon is 360°.

So, we have the equation as

x + 2x + (x - 1) + 3x + (x + 18) + (x + 10) = 360°

Collecting like terms, we have that

x + 2x + x + 3x + x + x - 1 + 18 + 10 = 360°

9x + 27° = 360°

Subtracting 27° from both sides of the equation, we have that

9x + 27° - 27° = 360° - 27°

9x + 0 = 333°

9x = 333°

Dividing both sides by 9, we have that

x = 333°/9

x = 37°

So, the value of x = 37°

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The range of the function for one day of work is 75 ≤ y ≤ 425. So, correct  option is B.

In mathematics, a function is a mathematical object that takes an input (or several inputs) and produces a unique output. It is a relationship between a set of inputs, called the domain, and a set of outputs, called the range.

Formally, a function f is defined by a set of ordered pairs (x, y) where x is an element of the domain, and y is an element of the range, and each element x in the domain is paired with a unique element y in the range. We write this as f(x) = y.

The linear function that models the daily cost of hiring an electrician can be written as:

y = 50x + 75

where x is the number of hours worked by the electrician and y is the cost in dollars.

Since the electrician works a maximum of 7 hours per day, the domain of the function is 0 ≤ x ≤ 7.

To find the range of the function, we can substitute the maximum and minimum values of x into the function and see what values of y we get:

When x = 0 (no hours worked), y = 50(0) + 75 = 75.

When x = 7 (maximum hours worked), y = 50(7) + 75 = 425.

Therefore, the range of the function for one day of work is:

75 ≤ y ≤ 425

So the answer is (B) 75 ≤ y ≤ 425.

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Answer:

Let's call the speed of Bug F v_F and the speed of Bug S v_S. Since both bugs started at 0, we can express their positions at any time t as:

Position of Bug F = 12 + v_F * t

Position of Bug S = 8 + v_S * t

To find out when F and S will be 100 units apart, we need to find the time t at which their positions differ by 100 units. In other words, we need to solve the following equation:

|12 + v_F * t - (8 + v_S * t)| = 100

We can simplify this equation by expanding the absolute value and rearranging the terms:

|4 + (v_F - v_S) * t| = 100

Now we can split this equation into two cases:

Case 1: 4 + (v_F - v_S) * t = 100

In this case, we have:

v_F - v_S > 0 (since Bug F is faster)

t = (100 - 4) / (v_F - v_S)

Case 2: 4 + (v_F - v_S) * t = -100

In this case, we have:

v_F - v_S < 0 (since Bug S is faster)

t = (-100 - 4) / (v_F - v_S)

Since we're only interested in positive values of t, we can discard the second case. Therefore, the time at which F and S will be 100 units apart is:

t = (100 - 4) / (v_F - v_S)

t = 96 / (v_F - v_S)

We don't know the values of v_F and v_S, but we can use the fact that Bug F is at 12 and Bug S is at 8, four seconds after they started. This gives us two equations:

12 = 4v_F + 0v_S

8 = 4v_S + 0v_F

Solving these equations for v_F and v_S, we get:

v_F = 3

v_S = 2

Substituting these values into the equation for t, we get:

t = 96 / (3 - 2)

t = 96

Therefore, F and S will be 100 units apart 96 seconds after they start.

Answer:

x = 0

Step-by-step explanation:

9x - 7 = -7

Add 7 to both sides.

9x - 7 + 7 = -7 + 7

9x = 0

Divide both sides by 9.

9x/9 = 0/9

x = 0

Answer:

x=0

Step-by-step explanation:

Answer: To solve the third-order differential equation y''' + y' = cot(x) by variation of parameters, we first need to find the solution to the associated homogeneous equation, which is:

y''' + y' = 0

The characteristic equation is r^3 + r = 0, which can be factored as r(r^2 + 1) = 0. This gives us the roots r = 0, r = i, and r = -i. Therefore, the general solution to the homogeneous equation is:

y_h = c1 + c2 cos(x) + c3 sin(x)

To find a particular solution to the non-homogeneous equation using variation of parameters, we assume that the solution has the form:

y_p = u1(x) + u2(x) cos(x) + u3(x) sin(x)

where u1, u2, and u3 are functions to be determined.

We can find the derivatives of y_p:

y'_p = u1'(x) + u2'(x) cos(x) - u2(x) sin(x) + u3'(x) sin(x) + u3(x) cos(x)

y''_p = u1''(x) + u2''(x) cos(x) - 2u2'(x) sin(x) - u2(x) cos(x) + u3''(x) sin(x) + 2u3'(x) cos(x) - u3(x) sin(x)

y'''_p = u1'''(x) + u2'''(x) cos(x) - 3u2''(x) sin(x) - 3u2'(x) cos(x) - u2(x) sin(x) + u3'''(x) sin(x) + 3u3''(x) cos(x) - 3u3'(x) sin(x)

Substituting these derivatives into the non-homogeneous equation, we get:

u1'''(x) + u2'''(x) cos(x) - 3u2''(x) sin(x) - 3u2'(x) cos(x) - u2(x) sin(x) + u3'''(x) sin(x) + 3u3''(x) cos(x) - 3u3'(x) sin(x) + u1'(x) + u2'(x) cos(x) - u2(x) sin(x) + u3'(x) sin(x) + u3(x) cos(x) = cot(x)

Grouping the terms with the same functions together, we get:

u1'''(x) + u1'(x) = 0

u2'''(x) cos(x) - 3u2''(x) sin(x) - u2(x) sin(x) + u2'(x) cos(x) + u2'(x) cos(x) = cot(x) cos(x)

u3'''(x) sin(x) + 3u3''(x) cos(x) - 3u3'(x) sin(x) + u3'(x) sin(x) + u3(x) cos(x) = cot(x) sin(x)

The first equation is a first-order differential equation, which can be solved by integrating both sides:

u1'(x) + u1(x) = c1

where c1 is a constant of integration. The solution to this equation is:

u1(x) = c1 + c2 e^(-x)

where c2 is another constant of integration.

Step-by-step explanation:

Answer:

∠w = 50°

∠y = 130°

Step-by-step explanation:

Angles ∠w and ∠y are supplementary angles, which means their sum is 180.

4x + 6 + 12x - 2 = 180

16x + 4 = 180

16x = 176

x = 11

To find the angle measures replace x with 11

∠w = 4x + 6

∠w = 4*11 + 6

∠w = 50°

Now, ∠y

∠y = 12x - 2

∠y = 12*11 - 2

∠y = 130°

Answer: 6 tablespoons

Answer: brainliest ?

The correct answer is: The data is dependent, because the hours of video streaming in January and July occurred in the same households.

This is because the data collected is related to the same households and not different people. The same households were monitored during both January and July, which means that any changes in the video streaming hours during those months are likely due to factors within those households, such as changes in habits, routines, or available free time.

If the data were independent, then it would mean that the hours of video streaming in January and July were not related or influenced by any factors from the other month. For example, if the data were collected from two completely different sets of households, then it would be considered independent because there would be no connection or influence between the households in January and July. However, this is not the case in the given scenario, so the correct answer is that the data is dependent.

Step-by-step explanation:

hope its help <:

Answer:

Thank you for your kind words and concern. I appreciate your offer to listen and support others. I think it's important for people to know that they can reach out for help when they're struggling and that there are resources available to them. And as for something that made me happy, I always appreciate it when people show kindness and empathy towards others, and when they're able to make a positive impact in the world.

Answer:hi tysm your literally the person everyone needs in there life and your so kind for that

Step-by-step explanation:

Today me and my friend went to the park with my brother and we got snowcones and played for a while then I got home and took a refreshing nap!

Answer: 3

Step-by-step explanation:

Answer:  3

Step-by-step explanation:

16/4=unit rate =4

1 in 4 hour

3 for 12 hours

Answer:

Below

Step-by-step explanation:

3.15 can be read as  3 and 15 hundredths =   3   15/100  = 3  3/20

Answer:  63/20

Step-by-step explanation:

Hence, 160 mL of oxygen is contained inside a 200 mL air tank using the same ratio as a 500 mL air tank.

Ratios contrast two figures by ordinarily dividing them. A/B would be your formula if you were trying to compare one piece of data (A) to some other data point (B). We are multiplying information A by data B, as this suggests. In the event that A and B are both 5, for example, your ratio would've been 5/10.

80% of 500 mL of oxygen if indeed the air tank contains 80% oxygen & 20% nitrogen, which is:

0.80 x 500 mL = 400 mL

This means that the air tank contains 400 mL of oxygen and 100 mL of nitrogen.

To find the amount of oxygen in a 200 mL air tank that contains the same ratio, we need to use proportions:

If 500 mL contains 400 mL of oxygen, then 1 mL contains 400/500 = 0.8 mL of oxygen.

Therefore, if 200 mL contains the same ratio of oxygen, it will contain:

0.8 mL x 200 mL = 160 mL

So, the amount of oxygen in a 200 mL air tank with the same ratio as the 500 mL air tank is 160 mL.

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