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victus00 [196]
3 years ago
13

A circle is graphed on a coordinate grid and then reflected across the y-axis. If the center of the original circle was located

at (x, y), which ordered pair represents the center of the new circle after the transformation?
A) (x, y)
B) (x, −y)
C) (−x, y)
D) (−x, −y)
Mathematics
2 answers:
lyudmila [28]3 years ago
8 0

Answer:

Option C. (-x,y)

Step-by-step explanation:

we know that

A point reflected across the y-axis has the following rule

(a,b)------> (-a,b)

so

If the center of the original circle was located at (x, y)

then

the center of the new circle after the transformation will be (-x,y)

Degger [83]3 years ago
4 0

Answer:

C.

Step-by-step explanation:

First, to see it better, you need to know that the y-axis can be referenced as "up or down" and the x-axis as "left or right".

Now, one ordered pair (x,y) after the reflection across the y-axis will be moved "from left to right" or "from right to left", then the y-value doesn't change and the x-value changes sign (to preserve the distance). Then, the the new center will be (-x,y). So, the answer is C.

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This set of points are connected to form a line segment. What is its length? A = (3,5) and B = (3,6) *
Colt1911 [192]

Answer:

d(A,B)=1

Step-by-step explanation:

To find distance between points A(xA,yA) and B(xB,yB), we use formula:

d(A,B)=(xB−xA)2+(yB−yA)2−−−−−−−−−−−−−−−−−−−√

In this example: xA=3 , yA=5 , xB=3 and yB=6 so:

d(A,B)d(A,B)d(A,B)=(3−3)2+(6−5)2−−−−−−−−−−−−−−−√=0+1−−−−√=1

5 0
3 years ago
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A sine function had an amplitude of 3, period of 6pi, horizontal shift of 3pi/2, &amp; vertical shift of -1.
Simora [160]

Answer: \bold{y=\dfrac{1}{2}}

<u>Step-by-step explanation:</u>

f(x) = A sin (Bx - C) + D

  • amplitude = |A|
  • period =\dfrac{2\pi}{B}
  • phase shift =\dfrac{C}{B}
  • vertical shift = D

<u>A</u>

amplitude of 3 is given so  3 = |A| → A = ± 3, since it is stated that this is a positive function, then A = 3

<u>B</u>

period of 6π is given so 6\pi=\dfrac{2\pi}{B}\quad \rightarrow \quad B=\dfrac{2\pi}{6\pi}\quad \rightarrow \quad B=\dfrac{1}{3}

<u>C</u>

\text{phase shift is given as}\ \dfrac{3\pi}{2}\ \text{so}\ \dfrac{3\pi}{2}=\dfrac{C}{\frac{1}{3}}\quad \rightarrow\quad \dfrac{(\frac{1}{3})3\pi}{2}=C\quad \rightarrow\quad \dfrac{\pi}{2}=C

<u>D</u>

vertical shift of -1 is given so -1 = D


Now, substitute the values of A, B, C, and D into the formula (above):

f(x) = 3\ sin \bigg(\dfrac{1}{3}x - \dfrac{\pi}{2}\bigg) - 1


Next, solve when x = 2π

f(2\pi) = 3\ sin \bigg(\dfrac{1}{3}(2\pi) - \dfrac{\pi}{2}\bigg) - 1

        = 3\ sin \bigg(\dfrac{2\pi}{3} - \dfrac{\pi}{2}\bigg) - 1

        = 3\ sin \bigg(\dfrac{4\pi}{6} - \dfrac{3\pi}{6}\bigg) - 1

        = 3\ sin \bigg(\dfrac{\pi}{6}\bigg) - 1

        = 3\ \bigg(\dfrac{1}{2}\bigg) - 1

        =\dfrac{3}{2}-\dfrac{2}{2}

        =\dfrac{1}{2}

6 0
3 years ago
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Answer:

(X - 3,  Y - 1).

3 0
3 years ago
PLEASE HELP ASAP WILL GIVE BRAINLIST!!
Mrac [35]

Answer: 18 units

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4 0
1 year ago
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