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just olya [345]
3 years ago
13

Write a rule to describe the function shown.

Mathematics
1 answer:
I am Lyosha [343]3 years ago
6 0

Answer:

y = -3x  

Step-by-step explanation:

(a) Slope

y = mx + b

Choose points (-3, 9) and (3, -9)

m = (-9 - 9)/(3 – (-3))

m = -18/(3 + 3)

m = -18/6

m = -3

=====

(b) y-intercept

y = mx +b

Choose point (0,0).

0 = 0 + b

b = 0

=====

(c) Equation of line

y = mx + b

y = -3x

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Can somebody solve this question?
lawyer [7]

Answer:

3 is the answer because it's value only add by the 3

8 0
2 years ago
Ben was walking in Manhattan at a constant speed. He went along 5th avenue from 4th to 98th street. At 3:00 he was on 15th stree
muminat

Answer: The starting time is 2:27 and finishing time is 7:09.    


Step-by-step explanation: It is given that Ben started from 4th street and he finished at 98th street.

Also, at 3:00, he was on 15th street and at 4:30, he was on 45th street.

That is, time taken to cover (45-15), i.e., 30 streets is 90 minutes,                           so the time taken to cover 1 street is 3 minutes.

Therefore, Ben covers distance from one street to second in 3 minutes. Since he started from 4th street, and there are 11 streets to cover between 4 and 15, so Ben's starting time was (3:00 - 3×11 min) = 2:27.

And his finishing time was (4:30 + 3×53 min) = 7:09.

Again, the equation that tells us on what street 'N' he was after time 'T' of his starting can be written as

N=4+\dfrac{T}{3.}

Thus, the starting and finishing time was 2:27 and 7:09 respectively.



3 0
3 years ago
A football is kicked from the ground and follows the graph of the function y = –t2 + 3t, where y represents height and t represe
vesna_86 [32]

Answer:

The height of the objects are the same after 2 seconds.

Step-by-step explanation:

In order to calculate at which time both objects have the same height we need to find the value of t that makes both equations equal. Therefore:

-t^2 + 3t = -t + 4\\t^2 - 3t - t + 4 = 0\\t^2 - 4t + 4 = 0\\t_{1,2} = \frac{-(-4) \pm \sqrt{(-4)^2 - 4*1*4}}{2*1} = \frac{4 \pm \sqrt{16 - 16}}{2}\\t_{1,2} = \frac{4}{2} = 2

The height of the objects are the same after 2 seconds.

8 0
3 years ago
Solve using the pathagorean theorem​
REY [17]

Answer:

b=45^1/2

Step-by-step explanation:

a^2+b^2=c^2\\6^2+b^2=9^2\\b^2=81-36\\b^2=45\\b=\sqrt{45}

4 0
2 years ago
x2 + y2 − 4x + 12y − 20 = 0 (x − 6)2 + (y − 4)2 = 56 x2 + y2 + 6x − 8y − 10 = 0 (x − 2)2 + (y + 6)2 = 60 3x2 + 3y2 + 12x + 18y −
snow_tiger [21]
For this case, what we must do is fill squares in all the expressions until we find the correct result.
 We have then:
 
 x2 + y2 − 4x + 12y − 20 = 0 x2 + y2  − 4x + 12y = 20
 x2  − 4x + y2 + 12y = 20
 x2  − 4x + (12/2)^2 + y2 + 12y  + (-4/2)^2 = 20 + (12/2)^2 + (-4/2)^2
 x2  − 4x + (6)^2 + y2 + 12y  + (-2)^2 = 20 + (6)^2 + (-2)^2
 x2  − 4x + 36 + y2 + 12y  + 4 = 20 + 36 + 4
 (x − 2)2 + (y + 6)2 = 60 

 
3x2 + 3y2 + 12x + 18y − 15 = 0 
 
x2 + y2 + 4x + 6y − 5 = 0 
 x2 + y2 + 4x + 6y  = 5 
 x2  + 4x + (4/2)^2 + y2 + 6y + (6/2)^2 = 5 + (4/2)^2 + (6/2)^2 
 x2  + 4x + (2)^2 + y2 + 6y + (3)^2 = 5 + (2)^2 + (3)^2 
 x2  + 4x + 4 + y2 + 6y + 9 = 5 + 4 + 9 
 (x + 2)2 + (y + 3)2 = 18 

 2x2 + 2y2 − 24x − 16y − 8 = 0
 x2 + y2 − 12x − 8y − 4 = 0
 x2 + y2 − 12x − 8y = 4 
 x2 − 12x + (-12/2)^2 + y2 − 8y + (-8/2)^2 = 4 + (-12/2)^2 + (-8/2)^2
 x2 − 12x + (-6)^2 + y2 − 8y + (-4)^2 = 4 + (-6)^2 + (-4)^2
 x2 − 12x + 36 + y2 − 8y + 16 = 4 + 36 + 16
 (x − 6)2 + (y − 4)2 = 56 

 x2 + y2 + 2x − 12y − 9 = 0
 x2 + y2  + 2x - 12y = 9
 x2  + 2x + y2 - 12y = 9
 x2  + 2x + (2/2)^2 + y2 - 12y  + (-12/2)^2 = 9 + (2/2)^2 + (-12/2)^2
 x2  + 2x + (1)^2 + y2 - 12y  + (-6)^2 = 9 + (1)^2 + (-6)^2
 x2  + 2x + 1 + y2 - 12y  + 36 = 9 + 1 + 36
 (x + 1)2 + (y − 6)2 = 46
7 0
4 years ago
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