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alexdok [17]
1 year ago
8

Max is trying to prove to his friend that two reflections, one across the x-axis and another across the y-axis, will not result

in a reflection across the line y = x for a pre-image in quadrant II. His friend Josiah is trying to prove that a reflection across the x-axis followed by a reflection across the y-axis will result in a reflection across the line y = x for a pre-image in quadrant II. Which student is correct, and which statements below will help him prove his conjecture? Check all that apply. Max is correct. O Josiah is correct. Taking the result from the first reflection (x, y) and applying the second mapping rule will result in (-X, -y), not (y x), which reflecting across the line y = x should give. If one reflects a figure first across the x-axis from quadrant II then reflects across the y-axis from quadrant II, the image will end up in quadrant IV. O A figure that is reflected from quadrant II to quadrant IV across the line y = x will have the coordinates of (-y).​
Mathematics
2 answers:
shtirl [24]1 year ago
7 0

Answer:

Max is correct

Taking the result from the first reflection (x, –y) and applying the second mapping rule will result in (–x, –y), not (y, x), which reflecting across the line y = x should give.

If one reflects a figure first across the x-axis from quadrant II then reflects across the y-axis from quadrant III, the image will end up in quadrant IV.

Step-by-step explanation

took the test on edge hope this helps

Leni [432]1 year ago
7 0

Answer:

1, 3, 4

TRUST ME I GOT IT RIGHT THE OTHER GUY IS WRONG

Step-by-step explanation:

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A motorcycle has an initial speed of u m/s. It accelerates to a speed of 1.2u in 10 seconds. The motorcycle then travels at a co
kolbaska11 [484]

Answer:

Step-by-step explanation:

A motorcycle has an initial speed of u m/s. It accelerates to a speed of 1.2u in 10 seconds

V = U + at

1.2u  = u  + a*10

=> a = 0.02u

S= ut + (1/2)at²

Distance in 1st 10 secs

S = u(10) + (1/2)(0.02)(10)²

=> S = 10u + u

=>  S = 11u  m

Constant speed 1.2u for 15 secs

S = 1.2u * 15

=> S = 18u m

Total Distance Covered d  = 11u + 18u = 29u  m

6 0
1 year ago
Would apreciate if someone helped me..
PtichkaEL [24]

Answer:

a). x = 11

b). m∠DMC = 39°

c). m∠MAD = 66°

d). m∠ADM = 36°

e). m∠ADC = 18°

Step-by-step explanation:

a). In the figure attached,

m∠AMC = 3x + 6

and m∠DMC = 6x - 49

Since "in-center" of a triangle is a points where the bisectors of internal angles meet.

Therefore, MC is the angle bisector of angle AMD.

and m∠AMC ≅ m∠DMC

3x + 6 = 8x - 49

8x - 3x = 49 + 6

5x = 55

x = 11

b). m∠DMC = 8x - 49

                   = (8 × 11) - 49

                   = 88 - 49

                   = 39°

c). m∠MAD = 2(m∠DAC)

                   = 2(30)°

                   = 60°

d). Since, m∠AMD + m∠ADM + m∠MAD = 180°

    2(39)° + m∠ADM + 66° = 180°

    78° + m∠ADM + 66° = 180°

    m∠ADM = 180° - 144°

                   = 36°

e). m∠ADC = \frac{1}{2}(m\angle ADM)

                   = \frac{1}{2}(36)

                   = 18°

5 0
1 year ago
Prompt:
Vadim26 [7]

Answer:

Her answer is wrong because she the object cannot hit the ground at negative seconds. She could’ve have used other methods because she used the quadratic formula. The advantages is that it works for every situation. The disadvantages is that it takes longer. She should’ve used a different method.

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
If X1 = 440, Y1 = 220, and X2 = 360, what’s the value of Y2
sergeinik [125]

Answer:

140 is the answer for y2 because x1 is 80 more than x2 so you would subtract y2 by 80

8 0
1 year ago
Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.950.950, point, 95 pro
bearhunter [10]

The question is incomplete. Here is the complete question:

Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.95 probability that he will hit it. One day, Samir decides to attempt to hit  10 such targets in a row.

Assuming that Samir is equally likely to hit each of the 10 targets, what is the probability that he will miss at least one of them?

Answer:

40.13%

Step-by-step explanation:

Let 'A' be the event of not missing a target in 10 attempts.

Therefore, the complement of event 'A' is \overline A=\textrm{Missing a target at least once}

Now, Samir is equally likely to hit each of the 10 targets. Therefore, probability of hitting each target each time is same and equal to 0.95.

Now, P(A)=0.95^{10}=0.5987

We know that the sum of probability of an event and its complement is 1.

So, P(A)+P(\overline A)=1\\\\P(\overline A)=1-P(A)\\\\P(\overline A)=1-0.5987\\\\P(\overline A)=0.4013=40.13\%

Therefore, the probability of missing a target at least once in 10 attempts is 40.13%.

6 0
1 year ago
Read 2 more answers
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