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TiliK225 [7]
3 years ago
11

I need help on number 2 and 3 please explain how to do this thanks!

Mathematics
1 answer:
Dominik [7]3 years ago
8 0
2. The answer is F. 15

3. The answer is C. 12
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A(9,2) B(1,6)<br> if (2,k) is equidistant from A and B, find the value of k.
Pani-rosa [81]
P is the point (2,k)
PA = PB
PA = √(49 + (2-k)²) and PB = √(1 + (6 - k)²)
√(49 + (2-k)²) = √(1 + (6 - k)²) => (49 + (2-k)² = (1 + (6 - k)²
=> 49 + 4 - 4k + k² = 1 + 36 - 12k + k² => 8k = 37 - 53 = -16 => k = -2

5 0
3 years ago
Screenshot of the question
tamaranim1 [39]

9514 1404 393

Answer:

  x = 1, x = 7

Step-by-step explanation:

You can see from the graph that the x-intercepts of f(x) are ...

  0 = f(-3)

  0 = f(3)

To find the corresponding values of x for f(x-4), we can solve ...

  0 = f(x -4)

  x -4 = -3   ⇒   x = 1

  x -4 = 3   ⇒   x = 7

The x-intercepts of the function after translation 4 units right are ...

  x = 1, x = 7

__

Your sketch will be the same curve moved 4 units to the right. (Add 4 to every x-value shown.)

3 0
3 years ago
Pls say hard very hard​
Simora [160]

(a) From the histogram, you can see that there are 2 students with scores between 50 and 60; 3 between 60 and 70; 7 between 70 and 80; 9 between 80 and 90; and 1 between 90 and 100. So there are a total of 2 + 3 + 7 + 9 + 1 = 22 students.

(b) This is entirely up to whoever constructed the histogram to begin with... It's ambiguous as to which of the groups contains students with a score of exactly 60 - are they placed in the 50-60 group, or in the 60-70 group?

On the other hand, if a student gets a score of 100, then they would certainly be put in the 90-100 group. So for the sake of consistency, you should probably assume that the groups are assigned as follows:

50 ≤ score ≤ 60   ==>   50-60

60 < score ≤ 70   ==>   60-70

70 < score ≤ 80   ==>   70-80

80 < score ≤ 90   ==>   80-90

90 < score ≤ 100   ==>   90-100

Then a student who scored a 60 should be added to the 50-60 category.

8 0
3 years ago
I need help please and thank you
sergeinik [125]
Try googling it , sorry
8 0
3 years ago
Please Help , Thank You !
kodGreya [7K]
One, Three and Four are right answers. 
7 0
4 years ago
Read 2 more answers
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