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Minchanka [31]
3 years ago
6

Solve for x and y: x+y=5 x−y=a PLEASE I NEED MY ANSWER TILL TOMORROW MORNING!

Mathematics
2 answers:
Daniel [21]3 years ago
8 0

Answer:

x=5/2 + a/2,  y= 5/2 - a/2

Step-by-step explanation:

We can solve this system using addition of the equations

x+y=5

x-y=a

2x  =(5+a)

x=(5+a)/2= 5/2 +a/2

x=5/2 +a/2

x-y=a

y = x - a , x=5/2 + a/2

y= 5/2 + a/2-a

y= 5/2 - a/2

Read more on Brainly.com - brainly.com/question/12520369#readmoreStep-by-step explanation:

KengaRu [80]3 years ago
7 0

Answer:

x=\frac{a+5}{2}  y=5-\frac{a+5}{2}  

Step-by-step explanation:

x+y=5

y=5-x

x-y=a

x-(5-x)=a

2x-5=a

2x=a+5

x=\frac{a+5}{2}  

x+y=5

\frac{a+5}{2}  +y=5

y=5-\frac{a+5}{2}  

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You are arguing over a cell phone while trailing an unmarked police car by 25 m; both your car and the police car are traveling
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Answer:

  (a) 15 m

  (b) ground speed: 26.14 m/s (94.1 km/h); relative speed 12 m/s

Step-by-step explanation:

We can choose the frame of reference to be the trailing car. We can start counting time from the point the lead car begins deceleration. Then its position (in meters) relative to the trailing car is ...

  d(t) = 25 - 5/2t² . . . . . . where 25 m is the initial distance and -5 is the acceleration in m/s².

(a) Then the distance between cars at t=2 is ...

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(b) The <em>relative velocity</em> at 2.4 seconds is ...

  d'(t) = -5t

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We need to do a little more work to find the ground speed of the trailing car when the cars bump.

The distance between the cars at t=2.4 seconds is ...

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At the closure speed of 12 m/s, it will take an additional ...

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to close the gap. The speed of the trailing car at that point in time will be the original speed less the deceleration for 0.8833 seconds:

  (110 km/h)·(1/3.6 (m/s)/(km/h))-(5 m/s²)(53/60 s)

     = 26 5/36 m/s = 94.1 km/h

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<em>Comment on following distance</em>

To avoid collision, the trailing car must be trailing by at least the distance it covers in 2.4 seconds, about 73 1/3 meters.

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3 years ago
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