find the curl of the vector field f. f(x, y, z) = (6y − z)i + ezj + xyzk

Answer: I think that you will have to times each side by 2 that whould equal 16.

Step-by-step explanation:

y = -3x - 4 is the figure's reflection, which is y = -3x + 2. An object with only one dimension, a line has length but no width .

A line is a straight, thickness-free, one-dimensional object in geometry that extends indefinitely in both directions. An object with only one dimension, a line has length but no width. A line is made up of several points that are endlessly stretched in the opposing directions. Collinear points are described as two points that are on the same line. An endlessly long, two-directional line is referred to as a line. It just has one, and that is length. Collinear points are those that are situated along the same path. A line is drawn with an arrowhead as it is defined by two points as illustrated below.

given

We know that the line y = 3x + 2y is reflected in the line y = -1

Observing the coordinate system, we conclude that the lines y = 3x + 2 and y=−1 intersect at the point whose coordinates are (-1,-1).

The line y = 3x + 2y intersect y-axis at the point whose coordinates are (0,-4).

So, the image must pass trough these two points (-1,-1) and (0,-4).

Now, we will substitute these two points in the equation of the line.

[tex]\frac{y - y2}{y2 - y1} = \frac{x - x2}{x2 - x1}[/tex]

x1 = -1

x2 = 0

y1 = -1

y2 = -4

[tex]\frac{y - ( -4 )}{-4 - ( -1 )} = \frac{x - 0}{0 - ( -1 )}[/tex]

[tex]\frac{y + 4}{3} = \frac{x}{1}[/tex]

y = -3x - 4

y = -3x - 4 is the figure's reflection, which is y = -3x + 2 .

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Answer:

2

Step-by-step explanation:

Point c is at (1, 2)

point d is at (2, 4)

rate of change formula

(f(d) - f(c))/2 - 1

f(2) = 4

f(1) = 2

4-2/1 = 2

Explore all sim

In the complex number 4 + 2i, 4 is the real part. In the complex number 4 + 2i, 2 is the imaginary part.

Complex numbers are those that are represented as a+ib, where a and b are actual numbers and i is an imaginary number termed a "iota."

Imaginary numbers are those that don't exist in reality. An imaginary number produces a negative value when squared. It is shown as Im (). Example: The numbers 2, 7, and 11 are all fictitious.

The two parts of the complex number are called the real part and the imaginary part. The imaginary part is identified by its multiplier of i.

In the given number, the 2 is multiplied by i, so 2 is the imaginary part. The other part, 4, is the real part.

Therefore, In the complex number 4 + 2i, 4 is the real part. In the complex number 4 + 2i, 2 is the imaginary part.

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In the complex number 4+2i, the real part is 4 and the imaginary part is 2. In the complex number 4+2i, 4 is the real part and 2 is the imaginary part.

In general, a complex number can be written as a + bi, where a is the real part and b is the imaginary part. For example, in the complex number 3 + 4i, the real part is 3 and the imaginary part is 4.

The real part of a complex number is the coefficient of the real number (in this case, 3) in the complex number. The imaginary part of a complex number is the coefficient of the imaginary unit i in the complex number.

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Answer:

  y = -5/3x -7

Step-by-step explanation:

You want the slope-intercept equation of the line that goes through points (-3, -2) and (0, -7).

The slope is given by the formula ...

  m = (y2 -y1)/(x2 -x1)

  m = (-7 -(-2))/(0 -(-3)) = -5/3 . . . . . . use the given point coordinates

The y-intercept is the point on the y-axis, where the y-value is -7.

The slope-intercept form of the equation is ...

  y = mx +b . . . . . . . line with slope m and y-intercept b

  y = -5/3x -7 . . . . . . line with slope -5/3 and y-intercept -7

The average value of function f(x,y,z)=x+z^2 is 3.

To find the average value of f(x, y, z) = x + z^2 on the truncated cone z^2 = x^2 + y^2, with 1 ≤ z ≤ 16, we need to evaluate the integral of f(x, y, z) over the region defined by the truncated cone, and divide by the volume of that region.

The volume of the truncated cone is given by 1/3 * π * (R^2 + r^2 + R*r), where R and r are the radii of the top and bottom circles of the cone respectively. In this case R = 16 and r = 1, so the volume is 1/3 * π * (256 + 1 + 16) = 85.33...

The integral of f(x, y, z) over the region defined by the truncated cone can be found by converting the problem to cylindrical coordinates and then evaluating the integral.

After this, The solution will be

∫∫∫ (x + z^2) dV

= ∫∫∫ (rcos(θ) + z^2) rdzdθdr

= ∫∫∫ (rcos(θ) + r^2) dzdθdr

= ∫∫ (r^2cos(θ) + r^3/3) dθdr

= (r^3/3) * ∫∫ cos(θ) dθdr

after evaluating this, the final answer will be

(1/3) * (r^3) * (∫ cos(θ) dθ) * (∫ dr)

= (1/3) * (1^3) * (sin(θ)) * (r^2)

= (1/3) * (1^3) * (sin(θ)) * (16^2 - 1^2)

= (1/3) * (1) * (sin(θ)) * (255)

= (255/3) * (sin(θ))

and the average value is (255/3) * (sin(θ)) / 85.33... = 3

Therefore, The average value of function f(x,y,z)=x+z^2 is 3.

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The equation that can be used to find the number of snowballs Jeffrey and his friends make after lunch is = 153 snowballs - 85 snowballs

The number of snowballs that Jeffrey and his friends made after lunch, can be found by the formula :

= Number of total snowballs made - Number of snowballs made before the boys went to lunch

Number of total snowballs made = 153 snowballs

Number of snowballs made before the boys went to lunch = 85 snowballs

The equation for the number of snowballs Jeffrey and his friends make after lunch is :
= 153 snowballs - 85 snowballs

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Answer:

28

Step-by-step explanation:

step 1: put into fraction form (but it's a ratio)

[tex]\frac{6s}{7b} =\frac{24s}{Xb}[/tex] -> ( s = soccer ) (b = basketball)

step 2: find out the ratio between 6 soccer balls and 24 soccer balls

(the difference is 24 divided by 6, with is 4)

step 3: multiply the basketballs by that ratio

(7b * 4 = 28b)

Therefore, the answer is 28 basketballs.

The correct option regarding which bus has the least spread among the travel times is given as follows:

Bus 14, with an IQR of 6.

First we consider the dot plot, which shows the number of times that each observation appears in the data-set.

Then we consider the interquartile range, which gives the difference between the third quartile and the first quartile of the data-set.

The interquartile range is a better measure of spread compared to the range of a data-set, as it does not consider outliers.

For groups of 15 students, we have that:

The quartiles for Bus 14 are given as follows:

Hence the IQR is of:

IQR = Q3 - Q1 = 18 - 12 = 6.

The quartiles for Bus 18 are given as follows:

Hence the IQR is of:

IQR = Q3 - Q1 = 16 - 9 = 7.

Hence Bus 14 is the more consistent bus, due to the lower IQR.

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Answer: i dont know  why my previous question was deleted for being unclear but lets forget it, The correct answer one be option 2 or B.

Step-by-step explanation:

i had got it right on the test

The area of square will be;

⇒ 16/81 cm²

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;

The width of square = 4/9 centimeters

Now,

We know that;

The area of square = Side × Side

Here, The width of square = 4/9 centimeters

Hence, The area of square with side 4/9 cm is,

⇒ The area of square = Side × Side

⇒ The area of square = 4/9 × 4/9

⇒ The area of square = 16/81 cm²

Thus, We get;

⇒ The area of square = 16/81 cm

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The iterated integral evaluates to zero when converted to polar coordinates.

The iterated integral can be converted to polar coordinates using the following equation:

0 0 4x2y dx dy = 2π ∫0 a r dr dθ

This can be further simplified by integrating with respect to r:

2π∫0 a (a2−r2) dr

The integral can now be evaluated by calculating the definite integral between 0 and a:

2π [a3/3 - r3/3] |0 a

The evaluated integral is then simplified to:

2π[a3/3 - a3/3]

Finally, the evaluated integral is multiplied by 2π to give the final result:

2π(0)

Therefore, the answer is 0.

The iterated integral can be converted to polar coordinates using the equation

0 0 4x2y dx dy = 2π ∫0 a r dr dθ.

This can then be further simplified by integrating with respect to r, resulting in

2π∫0 a (a2−r2) dr.

The integral can be evaluated by calculating the definite integral between 0 and a, which is

2π [a3/3 - r3/3] |0 a.

This evaluated integral is simplified to

2π[a3/3 - a3/3]

and is then multiplied by 2π to give the final result; 2π(0). Therefore, the answer is 0.

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Answer:

Step-by-step explanation:

300

Answer:

[tex]21[/tex]

Step-by-step explanation:

[tex]\textrm{If y varies directly with x we can write this relationship as}[/tex]

[tex]y = k \cdot x[/tex]

[tex]\textrm{k is referred to as the constant of proportionality.}[/tex]

[tex]\textrm{Substituting these values into the above equation we get}[/tex]

[tex]3 = k \cdot 2[/tex]

[tex]So\; k = \dfrac{3}{2}[/tex]

[tex]\mathrm{When }[/tex] [tex]x = 14,[/tex]

[tex]y = \dfrac{3}{2} \cdot 14[/tex]

[tex]y = 21[/tex]

The solution of the initial value problem: y" + 164 = 5(t – 2), y(0) = 3, y'(0) = 8, by using the Laplace transform is equals to 3/s - 8/s² - 166/s³ + 5/s⁴ .

We have, a intial value problem as present below,

y" + 164 = 5(t – 2) ---(1)

and y(0) = 3, y'(0) = 8

We have to solve (1) using laplace transform. Taking Laplace Transform both sides in (1), L[y" + 164] = L[5(t – 2)]

Since, Laplace transform is linear nature,

L [ y" ] + L [ 164] = L[5(t – 2)]

Now, determine the Laplace Transform of left hand side, L [ y" ] + 164 L [1]

= s²L [ y(t)] - s y(0) + y'(0) + 164L [ 1]

= s²L [ y(t)] - 3s + 8 + 164/s

( since, y(0) = 3, y'(0) = 8 , L(1) = 1/s )

Consider right hand side, L[5(t – 2)]

= L( 5t ) - L(2) = 5L(t) - 2 L(1)

= 5 (1!/s²) - 2/s = 5/s² - 2/s

Now, s²L [ y(t)] - 3s + 8 + 164/s = 5/s² - 2/s

=> s²L [ y(t)] = 3s - 8 - 164/s + 5/s² - 2/s

=> s²L[y(t) ] = 5/s² + 3s - 166/s - 8

=> L[ y(t)] = 5/s⁴ + 3/s - 166/s³ - 8/s²

=> L[ y(t)] = 3/s - 8/s² - 166/s³ + 5/s⁴

So, the Laplace Transform of Initial value problem (1), Y(s) is 3/s - 8/s² - 166/s³ + 5/s⁴

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The  McCandless method has four steps divide the stage, assign lights, use color and fill the empty spaces.

Setting up the McCandless method for a thrust stage involves several steps.

To light a thrust stage using the McCandless method, divide the space into separately illuminated sections as the first step. The size of the stage will determine how you do it, but there should be enough areas for uniform lighting throughout.

Assign lights to each part of your stage once you've divided it into lighting segments. Three lights per section is the traditional norm, although you should use as many or as few as necessary to provide a consistent effect.

Use color: After placing your lights in each region, you may use gels to balance out their hues. To balance out the shadows on the performers' faces, the McCandless technique recommends a mix of warm and cold hues.

Fill in the blanks: At this stage, your lighting setup should be clear enough for you to understand how it affects your actors and see any potential issues. Use smaller lights, such as Fresnels, to tone down any unnaturally lit areas and blend the performers' looks.

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The process of McCandless method is explained below.

According to the theory of "Into the Wild", the process of McCandless method calls for a combination of warm and cool colors to balance out the shadows on the actors' faces.

And here we know that for a double McCandless system, then you can achieve this by making two of the lights warm and the other two cool.

Here you can also have all your lights be similar, neutral colors.

Apart from this, we have the following steps to do it:

We have to choose the right type of visualization.

And then we have to declutter your visualization.

And then we have to focus your audience's attention.

And the final steps is to think Like a Designer.

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the centroid of the region is (x, y) = (10, 1000/3).

(x, y) = (10, 1000/3)

1. Set up the equation for the centroid formula: x = (1/A)∫y dx and y = (1/A)∫x dy

2. Find the area of the region: A = ∫(y2 - y1) dx

3. Calculate the integral: ∫y dx = x4/4 + C and ∫x dy = xy + C

4. Substitute the boundaries into the integrals and solve for C: x4/4 + C = 30x and xy + C = 0

5. Substitute the solutions for C in the centroid formula: x = (1/A)∫y dx = (1/A)(30x - x4/4) and y = (1/A)∫x dy = (1/A)(xy - 0)

6. Substitute the boundaries into the area equation and solve for A: A = ∫(y2 - y1) dx = ∫(30x - x4/4 - 0) dx = 30x2/2 - x5/5 + C

7. Substitute the solutions for C in A: A = 30x2/2 - x5/5 + C = 30(30)2/2 - (30)5/5 + C = 27000/2 - 27000 + C = 13500 + C

8. Substitute the solutions for C in the centroid formula and solve for x and y: x = (1/13500 + C)(30x - x4/4) and y = (1/13500 + C)(xy - 0)

9. Substitute the boundaries into the centroid formula and solve for x and y: x = 10 and y = 1000/3

Therefore, the centroid of the region is (x, y) = (10, 1000/3).

the complete question is :

Find the centroid of the region bounded by the given curves. y = x3, x + y = 30, y = 0 (x, y) = (10, 1000/3)

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So the equation x² + 2y² + 3z² + 2xy + 2yz - 2zx can be put into canonical form, (x + y + z)²/2.

The quadratic form you've provided can be put into canonical form by completing the square. The process involves a few steps:

Rearrange the terms so that all of the x, y, and z terms are on one side of the equation, with a constant term on the other side. In this case, the equation is already in this form.

Divide every term by 2, to make the next step more manageable. This gives us x²/2+ y²/2 + z²/2 + xy + yz - zx

For each variable, add and subtract the square of half of the coefficient of that variable's corresponding x,y,z coefficient. In this case, x²/2, y²/2, z²/2 are all squares so no action need to be done

Group like terms: (x²/2 + xy/2 + xz/2) + (y²/2 + yz/2 + yx/2) + (z²/2 + zx/2 + zy/2)

Each of the group in step 4 is a perfect square, and can be rewritten as (x/√2 + y/√2 + z/√2)² so that the equation becomes (x + y + z)²/2

So the equation x² + 2y² + 3z² + 2xy + 2yz - 2zx can be put into canonical form, (x + y + z)²/2. The conanical form is a useful representation of a quadratic form because it allows you to see the symmetry in the equation and also it allow us to see the level set of the form.

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For the given quadratic form equation x² + 2y² + 3z² + 2xy + 2yz - 2zx has the canonical form as [tex]\frac{ (x + y + z)^2}{2}[/tex]

Here we have to find the quadratic form you've provided can be put into canonical form by completing the square.

And the process involves a few steps,

First, we have to rearrange the terms so that all of the x, y, and z terms are on one side of the equation, with a constant term on the other side.

Now we have to divide every term by 2, to make the next step more manageable,

[tex]= > \frac{x^2}{2} + \frac{y^2}{2} + \frac{z^2}{2} + xy + yz - zx[/tex]

Now, we have to add and subtract the square of half of the coefficient of that variable's corresponding x,y,z coefficient, then we get the following,

[tex]= > \frac{x^2}{2} + \frac{y^2}{2} + \frac{z^2}{2} + xy + yz - zx[/tex]

And then we have to group like terms, then we get the equation like the following,

[tex]= > (\frac{x^2}{2} + \frac{xy}{2} + \frac{xz}{2} ) + (\frac{y^2}{2} + \frac{yz}{2} + \frac{yx}{2} ) + (\frac{z^2}{2} + \frac{zx}{2} + \frac{zy}{2} )[/tex]

In order to convert each of the group in step 4 is a perfect square, and then we have to rewritten as [tex](\frac{x}{\sqrt{2} } + \frac{y}{\sqrt{2} }+ \frac{z}{\sqrt{2} } )^2[/tex]

Then we get the canonical equation as

[tex]= > \frac{ (x + y + z)^2}{2}[/tex]

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P(both are greater than 15) is 3/16. The first tile is odd and the second tile is less than 25 is 7/20. The first tile is a multiple of 6 and the second tile is a multiple of 4 is 3/80 and The first tile is less than 15 and the second tile is even or greater than 25 is 21/50.

Given that,

Total number of outcomes = 20

1. P(both are greater than 15)   =  (5/20) (15/20)  =  (1/4) (3/4)  =  3/16

2. The first tile is odd and the second tile is less than 25  

so, (10/20) (14/20)  =  (1/2) (7/10) = 7/20

3. The first tile is a multiple of 6 and the second tile is a multiple of 4  

so, (3/20) ( 5/20) = (3/20) (1/4) = 3/80

4.  The first tile is less than 15 and the second tile is even or greater than 25

So, Number ( even or > 25)  = Number(even) + Number (.>25) - Number (even and > 25) = 10+5-3 = 12

Thus, (14/20) ( 12/20)  = (7/10) (3/5) = 21 / 50

Complete question:

Tiles numbered 1 through 20 are placed in a box. Tiles numbered 11 through 30 are placed in a second box. The first tile is randomly drawn from the first box. The second tile is randomly drawn from the second box. Find each probability.

1. P(both are greater than 15)

2. The first tile is odd and the second tile is less than 25.

3. The first tile is a multiple of 6 and the second tile is a multiple of 4.

4.  The first tile is less than 15 and the second tile is even or greater than 25.

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For a regular Hexagon , the formula for Area of hexagon is  [tex]\frac{3\sqrt{3} s^{2} }{2}[/tex] and Perimeter of Hexagon is  [tex]6s[/tex] .

What is a Regular Hexagon ?

A regular hexagon is defined as a closed shape polygon that  has six equal sides and six equal angles.

The area of the regular hexagon is the space that is enclosed by all of the six  sides .

let the side length of the regular polygon be = "s" ;

So , the Area will be denoted by = [tex]\frac{3\sqrt{3} s^{2} }{2}[/tex] ;

and the perimeter of regular hexagon is represented as = the sum of all six sides , that is "6s" .

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Answer:

[tex]\textsf{13.}\quad(x+1)^2-\dfrac{y^2}{3}=1[/tex]

[tex]\textsf{14.}\quad\dfrac{x^2}{16}-\dfrac{y^2}{65}=1[/tex]

[tex]\textsf{15.}\quad y^2-\dfrac{x^2}{9}=1[/tex]

[tex]\textsf{16.}\quad\dfrac{x^2}{36}-\dfrac{y^2}{64}=1[/tex]

Step-by-step explanation:

[tex]\boxed{\begin{minipage}{7.4 cm}\underline{Standard equation of a vertical hyperbola}\\\\$\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1$\\\\where:\\\phantom{ww}$\bullet$ $(h,k)$ is the center.\\ \phantom{ww}$\bullet$ $(h,k\pm a)$ are the vertices.\\\phantom{ww}$\bullet$ $(h,k\pm c)$ are the foci where $c^2=a^2+b^2$\\\phantom{ww}$\bullet$ $y =k\pm\left(\dfrac{a}{b}\right)(x-h)$ are the asymptotes.\\\end{minipage}}[/tex]

[tex]\boxed{\begin{minipage}{7.4 cm}\underline{Standard equation of a horizontal hyperbola}\\\\$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$\\\\where:\\\phantom{ww}$\bullet$ $(h,k)$ is the center.\\ \phantom{ww}$\bullet$ $(h\pm a,k)$ are the vertices.\\\phantom{ww}$\bullet$ $(h\pm c,k)$ are the foci where $c^2=a^2+b^2$\\\phantom{ww}$\bullet$ $y=k\pm\left(\dfrac{b}{a}\right)(x-h)$ are the asymptotes.\\\end{minipage}}[/tex]

Given:

As the y-values of the foci and vertices are the same, the hyperbola is horizontal (opening left and right).

The center (h, k) is the midpoint of the vertices.

Therefore, the center is (-1, 0) and so:

Use the formula for the vertices (h±a, k) to determine the value of a:

[tex]\begin{aligned}\implies h\pm a&=-2\\-1\pm a&=-2\\\pm a&=-1\end{aligned}[/tex]                [tex]\begin{aligned}\implies h\pm a&=0\\-1\pm a&=0\\\pm a&=1\end{aligned}[/tex]

Therefore:

Use the formula for the foci (h±c, k) to determine the value of c:

[tex]\begin{aligned}\implies h\pm c&=-3\\-1\pm c&=-3\\\pm c&=-2\end{aligned}[/tex]                [tex]\begin{aligned}\implies h\pm c &=1\\-1\pm c&=1\\\pm c &=2\end{aligned}[/tex]

Therefore:

To find b² use c² = a² + b² and the found values of a and c:

[tex]\begin{aligned}\implies c^2&=a^2+b^2\\2^2&=1^2+b^2\\4&=1+b^2\\b^2&=3\end{aligned}[/tex]

Substitute the found values of h, k, a² and b² into the formula to create an equation of the hyperbola:

[tex]\implies\dfrac{(x+1)^2}{1}-\dfrac{(y-0)^2}{3}=1[/tex]

[tex]\implies(x+1)^2-\dfrac{y^2}{3}=1[/tex]

Given:

As the y-values of the foci and vertices are the same, the hyperbola is horizontal.

The center (h, k) is the midpoint of the vertices.

Therefore, the center is (0, 0) and so:

Use the formula for the vertices (h±a, k) to determine the value of a:

[tex]\begin{aligned}\implies h\pm a&=\pm4\\0\pm a&=\pm4\\\pm a&=\pm4\end{aligned}[/tex]

Therefore:

Use the formula for the foci (h±c, k) to determine the value of c:

[tex]\begin{aligned}\implies h \pm c&=\pm9\\0\pm c&=\pm9\\\pm c&=\pm9\end{aligned}[/tex]

Therefore:

To find b² use c² = a² + b² and the found values of a and c:

[tex]\begin{aligned}\implies c^2&=a^2+b^2\\9^2&=4^2+b^2\\81&=16+b^2\\b^2&=65\end{aligned}[/tex]

Substitute the found values of h, k, a² and b² into the formula to create an equation of the hyperbola:

[tex]\implies\dfrac{(x-0)^2}{16}-\dfrac{(y-0)^2}{65}=1[/tex]

[tex]\implies\dfrac{x^2}{16}-\dfrac{y^2}{65}=1[/tex]

Given:

As the x-values of the vertices are the same, the hyperbola is vertical (opening up and down).

The center (h, k) is the midpoint of the vertices.

Therefore, the center is (0, 0) and so:

Use the formula for the vertices (h, k±a) to determine the value of a:

[tex]\begin{aligned}\implies k \pm a&=\pm1\\0\pm a&=\pm1\\\pm a&=\pm1\end{aligned}[/tex]

Therefore:

Use the formula for the asymptotes to determine the value of b:

[tex]\begin{aligned}\implies k\pm\left(\dfrac{a}{b}\right)(x-h)&=\pm\dfrac{1}{3}x\\\\0\pm \left(\dfrac{1}{b}\right)(x-0)&=\pm\dfrac{1}{3}x\\\\\pm\dfrac{1}{b}x&=\pm\dfrac{1}{3}x\\\\\pm b&=\pm3\end{aligned}[/tex]

Therefore:

Substitute the found values of h, k, a² and b² into the formula to create an equation of the hyperbola:

[tex]\implies\dfrac{(y-0)^2}{1}-\dfrac{(x-0)^2}{9}=1[/tex]

[tex]\implies y^2-\dfrac{x^2}{9}=1[/tex]

As the y-values of the vertices are the same, the hyperbola is horizontal.

The center (h, k) is the midpoint of the vertices.

Therefore, the center is (0, 0) and so:

Use the formula for the vertices (h±a, k) to determine the value of a:

[tex]\begin{aligned}\implies h \pm a&=\pm6\\0\pm a&=\pm6\\\pm a&=\pm6\end{aligned}[/tex]

Therefore:

Use the formula for the asymptotes to determine the value of b:

[tex]\begin{aligned}\implies k \pm\left(\dfrac{b}{a}\right)(x-h)&=\pm\dfrac{4}{3}x\\\\0\pm \left(\dfrac{b}{6}\right)(x-0)&=\pm\dfrac{4}{3}x\\\\\pm \dfrac{b}{6}x&=\pm\dfrac{4}{3}x\\\\ \pm b&=\pm8\end{aligned}[/tex]

Therefore:

Substitute the found values of h, k, a² and b² into the formula to create an equation of the hyperbola:

[tex]\implies \dfrac{x^2}{36}-\dfrac{y^2}{64}=1[/tex]

The value of x is 2.

An expression, rule, or law in mathematics that establishes the relationship between an independent variable and a dependent variable (the dependent variable). In mathematics and the sciences, functions are fundamental for constructing physical relationships.

Given

g(x) = 4x + 6.....g(x) = 14

14 = 4x + 6

subtract 6 both sides

14 - 6 = 4x

8 = 4x

divide by 4 both sides

8/4 = x

2 = x

Hence value of x is 2.

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Answer: 90 miles

Step-by-step explanation:

67+23=90

=

Two single receptacles can be installed in a two-gang box made up of two single-gang boxes ganged together.

A two gang-box is a square electrical box, also called a double-gang box, houses two devices. The two gang-box have a combination switch/outlet or a pair of switches/outlets inside that can control two lighting circuits.

A two-gang box made up of two single-gang boxes which are ganged together can install two single receptacles- receptacles refer to the openings in the gang box into which electronics can be plugged.

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The receiver should be placed 9 feet away from the base of the dish, along its axis of symmetry.

This was found by using the equation for a vertical parabola, y = (¹/₄a)(x - h)² + k,

where (h, k) are the coordinates of the center.

As the parabola is at the center, (h, k) = (0, 0) and the equation simplifies to y = ¹/₄ax² .

Using the given information that the dish is 12 feet across at its opening and 2 feet deep at its center, the coordinate of the given point is (12,4) and by substituting this information into the equation,

the value of a is found as 9 ft which represents the distance between the vertex and the focus.

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The approximation for f(−2.2) found by using the line tangent to the graph of F at x= -2 is f(-2.2) = 11

Linear approximation is the procedure of using derivatives to find the value of a function.

Given that F be a differentiable function such that f(-2) = -10 and f'(-2)= -5, and we desire f(2.2). The linear approximation is given as

f(x + Δx) = f(x) + f'(x)Δx

Since we require f(-2.2) = f(-2 - 0.2).

So,

So, substituting these into the equation, we have

f(-2 - 0.2) = f(-2) + f'(-2)(-0.2)

Given that

Substituting the values of the variables into the equation, we have that

f(-2 - 0.2) = f(-2) + f'(-2)(-0.2)

f(-2 - 0.2) = -10 + (5)(-0.2)

f(-2 - 0.2) = -10 - 1

f(-2.2) = -11

So, the approximation of f(-2.2) = -11

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The car wash made $600 since "there will be 8 teams" meaning 8 * $75.

The length of the car wash will be five and a half hours, with two 15-minute breaks for each team.

Thus, each team works for five hours.

Two automobiles will be washed per hour by each team.

Ten vehicles per squad, then.

[5*2]

Given that "there will be 8 teams," 80 vehicles will be washed (assumed to be in a continuous line of vehicles) [10*8].

The charge must be $600/80=$7.50 each car in order to make $600.

Check (extremely important): "each crew will wash two cars per hour," meaning that each team will earn $15 per hour; "the car wash will run five and a half hours," and "each team will take two 15-minute breaks." As a result, five times $15 equals $75 earned each team.

The car wash made $600 since "there will be 8 teams" meaning 8 * $75.

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Reporting frequencies of sample data, such as eye color, ethnicity, grade point average, and height is done by using (Option C.) descriptive statistics.

Descriptive statistics are used to summarize, organize, and describe sample data, such as eye color, ethnicity, grade point average, and height. These statistics are used to analyze data and make inferences about larger populations.

Descriptive statistics are used to calculate the mean, median, mode, and range of the data, as well as to provide visual representation of the data through charts and graphs. Descriptive statistics are one of the most commonly used forms of statistical analysis.

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Answer:

Step-by-step explanation:

The answer is d