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uranmaximum [27]
2 years ago
7

PLEASE HELP FAST! ITS EASY! WILL GIVE BRAINLIEST!

Mathematics
1 answer:
AveGali [126]2 years ago
8 0

Answer:

160 kilometers per hour

9560 kilometers per minute

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Shishir bought 4000 orange at 70 paisa each. But 400 of them were rotten. He sold 2000 oranges at 90 paisa each.If he plans to m
Ipatiy [6.2K]

Answer:

He needs to sell the rest of the oranges at 75 paisa each.

Step-by-step explanation:

Consider the given information that, Shishir bought 4000 orange at 70 paisa each.

Note: 1 rupees = 100 paisa

Thus, 70 paisa = 70/100 rupees = 0.70 rupees

Therefore, the cost price of 4000 oranges is:

4000×0.70 rupees = 2800 rupees

The selling price of 2000 oranges is:

2000×0.90 rupees = 1800 rupees

The number of oranges now Shishir have:

4000 - 2000 - 400 = 1600

He wants to make a profit of RS 200. Thus the selling price of 4000 oranges should be:

2800 rupees + 200 rupees = 3000 rupees

He earned 1800 rupees by selling 2000 oranges at 90 paisa. So, the remaining amount that he needs to make with 1600 oranges is:

3000 rupees - 1800 rupees = 1200 rupees

Therefore, the cost of one orange is:

1600 oranges = 1200 rupees

1 orange = 1200/1600 rupees

1 orange = 0.75 rupees

Hence, he needs to sell the rest of the oranges at 75 paisa each.

4 0
2 years ago
Which of these equations have no solution? Check all that apply. 2(x + 2) + 2 = 2(x + 3) + 1 2x + 3(x + 5) = 5(x – 3) 4(x + 3) =
o-na [289]

Answer:

  • 2(x + 2) + 2 = 2(x + 3) + 1
  • 2x + 3(x + 5) = 5(x – 3)
  • 5(x + 4) – x = 4(x + 5) – 1

Step-by-step explanation:

It can be easier to see the answer if you subtract the right side of the equation from both sides, then simplify.

1. 2(x + 2) + 2 = 2(x + 3) + 1

  2(x + 2) + 2 - (2(x + 3) + 1) = 0

  2x +4 +2 -2x -6 -1 = 0

  -1 = 0 . . . . no solution

__

2. 2x + 3(x + 5) = 5(x – 3)

  2x + 3(x + 5) - 5(x – 3) = 0

  2x +3x +15 -5x +15 = 0

  30 = 0 . . . . no solution

__

3. 4(x + 3) = x + 12

  4(x + 3) - (x + 12) = 0

  4x +12 -x -12 = 0

  3x = 0 . . . . one solution, x=0

__

4. 4 – (2x + 5) = (–4x – 2)

  4 – (2x + 5) - (–4x – 2) = 0

  4 -2x -5 +4x +2 = 0

  2x +1 = 0 . . . . one solution, x=-1/2

__

5. 5(x + 4) – x = 4(x + 5) – 1

  5(x + 4) – x - (4(x + 5) – 1) = 0

  5x +20 -x -4x -20 +1 = 0

  1 = 0 . . . . no solution

3 0
2 years ago
Read 2 more answers
Profit:p=r-c;solve for c
eduard
If you would like to solve p = r - c for c, you can do this using the following steps:

p = r - c         /+c
p + c = r - c + c
p + c = r       /-p
p + c - p = r - p
c = r - p

The correct result would be c = r - p.
5 0
2 years ago
Read 2 more answers
I've been stuck on this for so long and I have an exam soon, anybody who can help me :'( ?
san4es73 [151]
(i)  speed = distance / time
so time =  distance / speed
here we have

time t = 1080/x  hours

(ii) return flight  time  = 1080 / (x + 30)  hours

(a)  1080/x - 1080/(x + 30) = 1/2

Multiplying  through by the LCD 2x(x + 30) we get:-

1080*2(x + 30) - 2x*1080 = x(x+30)
2160x + 64800 - 2160x = x^2 + 30x
x^2 + 30x - 64800  = 0

(b)  factoring;  -64800 = 270 * -240  ans 270-240 = 30 so we have

(x + 270)(x - 240) = 0   so x = 240  ( we ignore the negative -270)

So the speed for outward journey is 240 km/hr

(c) time ffor outward flight = 1080 / 240 =  4 1/2  hours

(d) average speed for whole flight = distance / time
   Time for outward journey = 4.5 hours and time for  return journey = d / v
= 1080 / (240+30) =  4 hours
 Therefore the average speed for whole journey =  2160 / 8.5 = 254.1 km/hr
8 0
2 years ago
Solve the equation y′ + 3y = t + e^(−2t).
Leni [432]
Hello,

I am going to remember:

y'+3y=0==>y=C*e^(-3t)

y'=C'*e^(-3t)-3C*e^(-3t)

y'+3y=C'*e^(-3t)-3Ce^(-3t)+3C*e^(-3t)=C'*e^(-3t) = t+e^(-2t)
==>C'=(t+e^(-2t))/e^(-3t)=t*e^(3t)+e^t
==>C=e^t+t*e^(3t) /3-e^(3t)/9

==>y= (e^t+t*e^(3t)/3-e^(3t)/9)*e^(-3t)+D
==>y=e^(-2t)+t/3-1/9+D
==>y=e^(-2t)+t/3+k


4 0
2 years ago
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