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nasty-shy [4]
3 years ago
15

Jorge said that y-values would stay the same when you reflect a preimage across the line y=5 since the y-values stay the same wh

en you reflect a preimage acorss the y-axis. Is Jorge Correct?
Mathematics
1 answer:
OLga [1]3 years ago
3 0
No because y=5 is a horizontal line on a graph, in this scenario the x-values would stay the same
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HELP!! I only need the answer. Thank you!
zvonat [6]
The answer would be 6 3/8

Hope this helps c:
5 0
3 years ago
Maria stated that 5/6 is between 4/5 and 6/7
Vlada [557]

Answer:

It's true, if that's what your asking.

Step-by-step explanation:

5/6 = 0.833 to 3 sig.fig.

4/5 = 0.800 to 3 sig.fig.

6/7 = 0.857 to 3 sig.fig.

0.833 is more than 0.800 and less than 0.857.

3 0
3 years ago
I need help on this question
rewona [7]

x = -7

hope this helps!

8 0
3 years ago
66 athletes are running a race. A gold medal is to be given to the winner, a silver medal is to be
Dmitrij [34]

Answer:

274560

Step-by-step explanation:

We can "choose" one of the 66 athletes for the gold medal. That athlete can't win any other medals, so there are 65 athletes left.

We can then "choose" one of the 65 athletes left for the silver medal. That athlete can't win any other medals, so there are 64 athletes left.

We can then "choose" one of the 64 athletes left for the bronze medal.

That leaves us with 66 possible choices * 65 possible choices * 64 possible choices=274560 possible choices

8 0
2 years ago
Compare the light gathering power of an 8" primary mirror with a 6" primary mirror. The 8" mirror has how much light gathering p
tankabanditka [31]

Answer:

1.778 times more or 16/9 times more

Step-by-step explanation:

Given:

- Mirror 1: D_1 = 8''

- Mirror 2: D_2 = 6"

Find:

Compare the light gathering power of an 8" primary mirror with a 6" primary mirror. The 8" mirror has how much light gathering power?

Solution:

- The light gathering power of a mirror (LGP) is proportional to the Area of the objects:

                                           LGP ∝ A

- Whereas, Area is proportional to the squared of the diameter i.e an area of a circle:

                                           A ∝ D^2

- Hence,                              LGP ∝ D^2

- Now compare the two diameters given:

                                           LGP_1 ∝ (D_1)^2

                                           LGP ∝ (D_2)^2

- Take a ratio of both:

                           LGP_1/LGP_2 ∝ (D_1)^2 / (D_2)^2

- Plug in the values:

                               LGP_1/LGP_2 ∝ (8)^2 / (6)^2

- Compute:             LGP_1/LGP_2 ∝ 16/9 ≅ 1.778 times more

6 0
3 years ago
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