Find the area of the parallelogram with vertices at (−2,3),(−2,3), (6,14),(6,14), (−14,1),(−14,1), and (−6,12).(−6,12).

Answer:

A,B,D,E

Step-by-step explanation:

did it on edge

You can calculate the scale factor by dividing the coordinates of the image by the corresponding coordinates of the pre-image.

A scale factor is defined as the ratio between the scale of a given original object and a new object, which is its representation but of a different size (bigger or smaller).

Assume that the initial coordinates are (x, y) and that the dilated coordinates are (x', y').

The dilation is therefore:

(x, y) → (x', y')

Now, let's assume that the dilation factor is k.

Therefore:

x' = kx

y' = ky

To get the initial coordinates and the final ones and then substitute in any of the above two equations to get the value of k, which is the scale factor,

For example =

Assume an original point at (2,4) is dilated to coordinates (4,8) we need to find the dilation factor.

Assume the dilation coefficient is k.

(x, y) are (2,4) and (x', y') are (4,8)

Therefore:

x' = kx → 4 = k×2 → k = 2

or:

y' = ky → 8 = k×4  → k = 2

Based on the above, the dilation coefficient would be 2.

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Answer(s):

[tex]\displaystyle y = 4cos\:(4\theta \pm \pi) \\ y = -4cos\:4\theta[/tex]

Step-by-step explanation:

[tex]\displaystyle y = Acos(B\theta - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{\pm\frac{\pi}{4}} \hookrightarrow \frac{\pm\pi}{4} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{\frac{\pi}{2}} \hookrightarrow \frac{2}{4}\pi \\ Amplitude \hookrightarrow 4[/tex]

OR

[tex]\displaystyle y = -Acos(B\theta - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{\frac{\pi}{2}} \hookrightarrow \frac{2}{4}\pi \\ Amplitude \hookrightarrow 4[/tex]

You will need the above information to help you interpret the graph. First off, keep in mind that this cosine graph will have TWO equations because the curvature begins upward from [tex]\displaystyle [0, -4][/tex] instead of downward from [tex]\displaystyle [0, 4],[/tex] telling you that one equation will have a “negative” symbol inserted in the beginning of the equation. Before we go any further though, we must figure the period of the graph out. So, in this case, sinse you ONLY have a graph to wourk with, you MUST figure the period out by using wavelengths. So, looking at where the graph hits [tex]\displaystyle [0, -4],[/tex]from there to [tex]\displaystyle [-\frac{\pi}{2}, -4],[/tex] they are obviously [tex]\displaystyle \frac{\pi}{2}\:unit[/tex]apart, telling you that the period of the graph is [tex]\displaystyle \frac{\pi}{2}.[/tex] Now, as you can see, the photograph on the right displays the trigonometric graph of [tex]\displaystyle y = 4cos\:4\theta.[/tex] Now, if you look hard enough, you will see that both graphs are “mirror reflections” of one another, meaning you can figure the rest of the terms out one of two ways. The first way is to figure the appropriate C-term out that will make the graph horisontally shift and map onto the original cosine graph [photograph on the left], accourding to the horisontal shift formula above. Also, keep in mind that −C gives you the OPPOCITE TERMS OF WHAT THEY REALLY ARE, so you must be careful with your calculations. So, between the two photographs, we can tell that the rightward graph is shifted [tex]\displaystyle \frac{\pi}{4}\:unit[/tex] on both sides of the y-axis, which means that in order to match the original graph, we need to shift the graph back, which means the C-term will be both negative and positive; and by perfourming your calculations, you will arrive at [tex]\displaystyle \boxed{\pm\frac{\pi}{4}} = \frac{\pm\pi}{4}.[/tex]So, one equation of the cosine graph, accourding to the horisontal shift, is [tex]\displaystyle y = 4cos\:(4\theta \pm \pi).[/tex] Now that we got this out the way, we can focuss on finding the second equation. Another way is to write an equation with a “negative” symbol inserted in the beginning [like I mentioned earlier]. Now, sinse we are writing an equation with the negative, the graph will not have a horisontal shift; so, C will be zero. With this said, the second equation is [tex]\displaystyle y = -4cos\:4\theta.[/tex] Now, the amplitude is obvious to figure out because it is the A-term, but of cource, if you want to be certain it is the amplitude, look at the graph to see how low and high each crest extends beyond the midline. The midline is the centre of your graph, also known as the vertical shift, which in this case the centre is at [tex]\displaystyle y = 0,[/tex] in which each crest is extended four units beyond the midline, hence, your amplitude. So, no matter how far the graph shifts vertically, the midline will ALWAYS follow.

I am delighted to assist you at any time.

The slope of the tangent will be "[tex]10a-6a^2[/tex]". The equation at point (1, 8) is "[tex]y = 4x+4[/tex]" and at point (2,9) is "[tex]y = -4x+17[/tex]".

Given equation is:

(a)

The slope of the tangent to the curve point where [tex]x = a[/tex] is:

→ [tex]\frac{dy}{dx} = m = 10x-6x^2[/tex]

→              [tex]= 10a-6a^2[/tex]

(b)

The equation of tangent at point (1,8) will be:

→ Slope, [tex]m = 10-6[/tex]

                  [tex]= 4[/tex]

→ [tex]y-8=4(x-1)[/tex]

         [tex]y = 4x-4+8[/tex]

         [tex]y = 4x+4[/tex]

and,

The equation of tangent at point (2,9) will be:

→ Slope, [tex]m = 10\times 2-6\times 2^2[/tex]

                  [tex]= 20-24[/tex]

                  [tex]= -4[/tex]

→ [tex]y-9=-4(x-2)[/tex]

   [tex]y-9=-4x+8[/tex]

         [tex]y = -4x+17[/tex]

Thus the above answer is right.

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Answer:

B. 20 inches

Step-by-step explanation:

Remember that when dealing with ratios or unit rates you just have to do simple rules of three to calcuate what they are asking you, in this case you have that in the map for every 8 units in the length there will be 5 in the width, so how many will there be when the map is 32 inches in length?

[tex]\frac{8}{5}= \frac{32}{x}\\x=\frac{32*5}{8}\\ x=20[/tex]

So the width of a map with a length of 32 inches and a ratio of 8:5 will be 20 inches.

The required correct answer is not listed among the given options. The area of the circle is approximately 50.24 square inches.

The area of a circle can be calculated using the formula A = πr^2, where A is the area and r is the radius of the circle. Since the given information provides the diameter (twice the radius), we need to first find the radius by dividing the diameter by 2.

Given: Diameter = 8 inches

Radius = Diameter / 2 = 8 inches / 2 = 4 inches

Now we can calculate the area using the formula:

= πr² = π * (4 inches)²
≈ 3.14 * 16 square inches
≈ 50.24 square inches

Therefore, the correct answer is not listed among the given options. The area of the circle is approximately 50.24 square inches.

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Final answer:

The area of a circle with an 8-inch diameter is calculated using the formula A = πr². The correct area is around 50.27 square inches, not one of the provided options.

Explanation:

The question asks for the area of a circle with a diameter of 8 inches. To find the area, we use the formula A = πr², where π is approximately 3.14159, and r is the radius of the circle. Since the diameter is 8 inches, the radius (r) is half of that, which is 4 inches.

To calculate the area:

The area of the circle is approximately 50.27 square inches, which means none of the given options A, B, C, or D is correct.