A bus goes from town A to B in an exact time. If the bus goes at the rate of 50km/h, then it will arrive B 42min later than it had to and if it increases its speed 5.5/9 m/sec., it will arrive B 30min earlier. Find:A) The distance between the two towns;
B) The exact time that it takes to arrive town B
C) The speed of the bus(by schedule) for the exact time.

Answers

Answer 1

Answer: Let the speed of bus for the exact time = x km/h
the distance between the cities = y km
then the exact time would be, t hours = (y/x) hours

If the bus goes at the rate of 50km/h, then it will arrive B 42min later,
speed = 50 km/h
42 minutes = 42/60 hours = 7/10 hours = 0.7 hours
time taken = t+0.7
distance = speed×time
⇒ y = 50×(t+0.7)
⇒ y = 50t + 35 ---------------------(1)

it increases its speed 5.5/9 m/sec, it will arrive B 30min earlier.
5.5/9 m/s = (5.5/9)×(18/5) km/h = 2.2 km/h
30 minutes = 30/60 = 0.5 hour
speed = (x+2.2) km/h
time = (t - 0.5) hours

distance = speed×time
⇒ y = (x+2.2)×(t-0.5)
⇒ y = ((y/t) +2.2)×(t-0.5) (t = y/x)
⇒ y = y - 0.5 (y/t) + 2.2t - 1.1
⇒ 0.5 (y/t) - 2.2t + 1.1 = 0   (subtracting y from both sides)
⇒ (y/t) - 4.4t - 2.2 = 0    (dividing both sides by 0.5)
⇒ y - 4.4t² - 2.2t = 0    (multiplying both sides by t)
⇒ 50t + 35 - 4.4t² - 2.2 t = 0 (from equation 1)
⇒ -4.4t² + 35 + 47.8t = 0
⇒ 4.4t² - 47.8t - 35 = 0

solving the quadratic equation, we get t = 11.55 hours
y = 50t + 35 = 612.5 km
x = 612.5/11.55 = 53 km/h

A) 612.5 km
B) 11.55 hours
C) 53 km/h

Answer 2

Answer: The Logic Defined:

1 Minute=t, (a unit of time)

Time (By schedule)=nt, (n>0), nt=number of minutes

Metre(s)=m

Speed=s (in metres per minute), s=[distance in metres]/[time in minutes]

Distance=d (in metres), d=[speed in metres per minute]*[time in minutes]

---------------------------------------------

Statement (1):

"If the bus goes at the rate of 50km/h, then it will arrive B 42min later than it had to."

Conclusion 1:

$\frac { 50km }{ h } =\frac { 50,000m }{ 60t } =\frac { 2,500m }{ 3t } \n \n \therefore \quad \frac { 2,500m }{ 3t } =nt+42t\n \n \frac { 2,500m }{ 3t } =t\left( n+42 \right)$

$\n \n 2,500m=3{ t }^( 2 )\left( n+42 \right) \n \n m=\frac { 3{ t }^( 2 )\left( n+42 \right) }{ 2,500 }$

Statement (2):

"if it increases its speed 5.5/9 m/sec., it will arrive B 30min earlier."

Conclusion 2:

$\frac { 5.5m }{ 9\quad seconds } =\frac { 5.5m }{ \frac { 9 }{ 60 } t } =\frac { 110m }{ 3t }$

$\n \n \therefore \quad \frac { 110m }{ 3t } =nt-30t\n \n \frac { 110m }{ 3t } =t\left( n-30 \right) \n \n 110m=3{ t }^( 2 )\left( n-30 \right)$

$\n \n m=\frac { 3{ t }^( 2 )\left( n-30 \right) }{ 110 }$

Conclusion 3, because of conclusion 1 and 2:

$\frac { 3{ t }^( 2 )\left( n-30 \right) }{ 110 } =\frac { 3{ t }^( 2 )\left( n+42 \right) }{ 2,500 } \n \n 7,500{ t }^( 2 )\left( n-30 \right) =330{ t }^( 2 )\left( n+42 \right) \n \n 7,500\left( n-30 \right) =330\left( n+42 \right)$

$\n \n 7,500n-225,000=330n+13,860\n \n 7,500n-330n=13,860+225,000\n \n 7,170n=238,860\n \n n=\frac { 238,860 }{ 7,170 } \n \n \therefore \quad n=\frac { 7962 }{ 239 }$

Therefore,

$Time\quad by\quad schedule=\frac { 7962 }{ 239 } t\n \n Approx:\quad 33.3\quad mins$

Now we want to find the distance between the two towns, so we say that:

$d=\frac { 2,500m }{ 3t } \cdot \left( \frac { 7962 }{ 239 } t+42t \right) \n \n =\frac { 2,500m }{ 3t } \cdot \frac { 18,000 }{ 239 } t$

$\n \n =\frac { 45,000,000 }{ 717 } m\n \n Approx:\quad 62,761.5\quad metres\n \n In\quad km\quad (approx):\quad 62.761\quad km$

So now you want to know how fast the bus has to travel to get to its destination on time...

Use the formula: s=d/t

Therefore:

$s=\frac { \frac { 45,000,000 }{ 717 } m }{ \frac { 7962 }{ 239 } t } \n \n Approx:\quad 1,883.9\quad metres\quad per\quad minute$

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At an airshow, 8 skydivers were released from a plane in a circle formation. Each skydiver was connected to each of the other skydivers with a separate pieces of ribbon. How many pieces of ribbon were used in the skydiving act?

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The total number of pieces of ribbons that were used in the skydiving act is 28 ribbons

What are Combinations?

The number of ways of selecting r objects from n unlike objects is:

The equation for combination is

ⁿCₓ = n! / ( ( n - x )! x! )

where

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x = number of choosing objects from the set

Given data ,

The total number of skydivers in the plane n = 8 skydivers

Each skydiver was connected to each of the other skydivers with a separate pieces of ribbon

So , the number of choosing skydivers x = 2

Now , the total number of ribbons used is calculated by combination

So , the equation is

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Substitute the value of n and x in the equation , we get

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Answer:

Step-by-step explanation:

If a ribbon connects two skydivers, and there are 8 skydivers total, then we can do 8C2, which is equal to (8*7)/2, which is equal to 28. An alternate solution is to think of it this way. Skydiver 1 is connected to everyone else, which is 7 other people. Skydiver 2 is connected to everyone else but Skydiver 1, which is 6 other people. This goes on until Skydiver 8 is already connected with everyone. So we have 7+6+5+4+3+2+1+0, which is also equal to 28.

What is the value of 5^3i^9
-125i
-15i
15i
125i

Answers

Good morning ☕️

______

Answer:

125i

___________________

Step-by-step explanation:

5³× i⁹ = 125 × i¹ = 125i

By the way :

i⁹ = (i⁴)²× i¹ = (1)²×i = i.

Answer:

$125i$

Step-by-step explanation:

When we have potential expression that include imaginary numbers, we have to consider some basic results, because these imaginary potential expression are cyclical.

We know that:

$i=√(-1)$

So, elevating both members to a third power, we have:

$i^(3)=(√(-1) )^(3)=\sqrt{(-1)^(3) }=√(-1)=i$

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So, from the problem, we have:

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A+b = 30 a:b is equivalent to 2:3. What is a& b

Answers

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Final answer:

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Explanation:

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Since the ratio 'a:b' is equal to '2:3', we can consider 'a' to be '2x' and 'b' to be '3x'. Substituting this into the first equation, we get '2x + 3x = 30', which simplifies to '5x = 30'. Solving for 'x', we find that 'x = 6'.

Substituting 'x = 6' into the expressions for 'a' and 'b', we find that 'a = 2*6 = 12' and 'b = 3*6 = 18'. Therefore, the solutions to the equations are 'a = 12' and 'b = 18'.

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