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AleksAgata [21]
3 years ago
6

Sometimes two transformations, one performed after the other, have a nice description as a single transformation. For example, i

nstead of translating 2 units up followed by translating 3 units up, we could simply translate 5 units up. Instead of rotating 20 degrees counterclockwise around the origin followed by rotating 80 degrees clockwise around the origin, we could simply rotate 60 degrees clockwise around the origin.
Can you find a simple description of reflecting across the x-axis followed by reflecting across the y-axis?
Mathematics
1 answer:
almond37 [142]3 years ago
6 0

Answer:

We rotate the point counterclockwise by an angle θ = tan⁻¹ (y/x)

Step-by-step explanation:

If we have the point (x, y), its reflection along the x-axis (x,-y). The reflection of the point (x, -y) along the y - axis is (-x, -y). So, the angle between the initial point (x, y) and the final point (-x, -y) is θ = tan⁻¹ [(-y - y)/(- x -x)] = tan⁻¹ [(-2y/(-2x)] =  tan⁻¹(y/x).

So, the point (x, y) is rotated counterclockwise by an angle of  θ = tan⁻¹ (y/x) to perform both reflections.

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51 degrees 44 degrees 31 degrees 42 degrees 52 degrees 39 degrees and 55 degrees what are the average median and mode of these t
yarga [219]
Median = 44 mode= none
6 0
3 years ago
1. Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%282x%2B3%29%20%7D" id="TexFormula1" title="\frac{1}{x(2x+3) }" a
katovenus [111]

1. Let a and b be coefficients such that

\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}

Combining the fractions on the right gives

\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}

\implies 1 = (2a+b)x + 3a

\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23

so that

\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}

2. a. The given ODE is separable as

x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}

Using the result of part (1), integrating both sides gives

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C

Given that y = 1 when x = 1, we find

\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)

so the particular solution to the ODE is

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)

We can solve this explicitly for y :

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)

\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|

\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|

\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}

2. b. When x = 9, we get

y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}

8 0
3 years ago
The perimeter of a rectangle downtown courtyard is 174 feet. The width of the rectangle is twice the length. Find the length and
kirill [66]

Answer:

Step-by-step explanation:

L = Length

W = Width

W = 2L

   2W + 2L = 174

2(2L) + 2L = 174

            6L = 174

              L = 29 ft

W = 2(29) = 58 ft

5 0
3 years ago
the number of a certain company’s video stores can be approximated by the linear equation y=-264x+4682, where x is the number of
kogti [31]
This problem is incomplete, but I think I know the probable question. Normally, you should have been given the target number of the company's video stores, so that you can replace the value of y and solve for x. Once you solve for x, you add this to year 1990 then you can solve for the year at which you can reach your target value. Suppose y = 3500 stores.

3500 = -264x + 4682
3500 - 4682 = -264x
-1182 = -264x
x = 4.48 years

Technically, that is equal to 4 years. Hence, the year would be 1990+4 = 1994.
8 0
4 years ago
Ummm plz help I dont know ​
SVETLANKA909090 [29]

Answer:

8/45

Step-by-step explanation:

1x8 is 8 and 5x9 is 45

3 0
3 years ago
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