Mathematics: Post your doubts here!

[The Sarcastic Retard]

& Sure. Inbox them or post them here

Thanks for the help. I have posted here, check my post.

 

[Anum96]

Thanks for the help. I have posted here, check my post.

Q11. b
3 + I2x - 1I = x
Subtract 3 from both sides.
I2x - 1I = x - 3
1. 2x - 1 = x - 3 ; x = -2
2. 2x - 1 = -(x - 3) ; x = 4/3
Is it?
Now check if they satisfy the equation
4/3 wont satisfy so its an extraneous solution.
-2 will satisfy so thats your answer

 

[The Sarcastic Retard]

Q11. b
3 + I2x - 1I = x
Subtract 3 from both sides.
I2x - 1I = x - 3
1. 2x - 1 = x - 3 ; x = -2
2. 2x - 1 = -(x - 3) ; x = 4/3
Is it?
Now check if they satisfy the equation
4/3 wont satisfy so its an extraneous solution.
-2 will satisfy so thats your answer

Answer is no solution.

 

[Anum96]

B

Answer is no solution.

Thats because absolute value is always positive So just write a brief statement along with your answer
Look.
You have 4/3 thats extraneous forget about that.
Now you have, -2. put that inside the absolute value brackets. You will get -5 hence no solution

 

[Anum96]

Thanks for the help. I have posted here, check my post.

Q15 a
1. x < 1
2. x > 5/3

b.
5/3<x<-5/3
-1<x<1

Tell me if im right. Ill do the working.

 

[The Sarcastic Retard]

Q15 a
1. x < 1
2. x > 5/3

b.
5/3<x<-5/3
-1<x<1

Tell me if im right. Ill do the working.

How? Its correcct.

 

[Anum96]

How? Its correcct.

Make the RHS equal to zero.
[(x+1)/(x-1)] -4 < 0
Take LCM and solve. You will get [(-3x + 5)/(x - 1)] <0
Now take the numerator. Remember its less than zero which means the LHS should be a negative value.
So,
-3x + 5 <0 is positive for x<5/3 And negative for x>5/3 So x>5/3 Is your answer no.1
Second. Take the denominator x - 1 is positive for x>1 and negative for x<1 and so x<1 is your answer no.2


for part b.
(-x+1)/(-x-1)<4
Solve it the way you solved a part. But because of different signs you will get
x<-5/3
and x>-1

And because of absolute value,
Solve this as well.
(x+1)/(x-1)<4
You will get x<1
and x>5/3
So altogether it becomes. 5/3<x<-5/3
-1<x<1

 

[The Sarcastic Retard]

Make the RHS equal to zero.
[(x+1)/(x-1)] -4 < 0
Take LCM and solve. You will get [(-3x + 5)/(x - 1)] <0
Now take the numerator. Remember its less than zero which means the LHS should be a negative value.
So,
-3x + 5 <0 is positive for x<5/3 And negative for x>5/3 So x>5/3 Is your answer no.1
Second. Take the denominator x - 1 is positive for x>1 and negative for x<1 and so x<1 is your answer no.2


for part b.
(-x+1)/(-x-1)<4
Solve it the way you solved a part. But because of different signs you will get
x<-5/3
and x>-1

And because of absolute value,
Solve this as well.
(x+1)/(x-1)<4
You will get x<1
and x>5/3
So altogether it becomes. 5/3<x<-5/3
-1<x<1

Thanks.

 

[DESTROYER1198]

Page 28 of:
http://www.cie.org.uk/images/164759-2016-syllabus.pdf

thanks

 

[nehaoscar]

How to solve this normal distribution?
P(|Z|>2.4)
The answer us 0.0164
Please tell me which identities to use here...

 

[Anum96]

How to solve this normal distribution?
P(|Z|>2.4)
The answer us 0.0164
Please tell me which identities to use here...

Use standard normal distribution
Z ~ N(0,1)

 

[The Sarcastic Retard]

Q2 Physicist

 

[Anum96]

Q2 Physicist
View attachment 57981

Let distances be s1 and s2 (use s = ut + 1/2at^2)
s1 + s2 = 10 s1 = 0.25t^2 and s2 = 0.75t
s(total) = ut + 1/2 at^2
10 = 0.25t^2 + 0.75t (multiply the whole expression by 100 to make it simpler)
25t^2 + 75t - 1000 = 0
t^2 + 3t - 40 = 0
t^2 + 8t - 5t - 40 = 0
t(t+8) - 5(t+8) = 0
t = 5 & t = -8
t = 5 accepted. (Ans)

ii) v = u + at
v = o + 0.5(5)
v = 2.5 (Ans)

 

[The Sarcastic Retard]

Let distances be s1 and s2 (use s = ut + 1/2at^2)
s1 + s2 = 10 s1 = 0.25t^2 and s2 = 0.75t
s(total) = ut + 1/2 at^2
10 = 0.25t^2 + 0.75t (multiply the whole expression by 100 to make it simpler)
25t^2 + 75t - 1000 = 0
t^2 + 3t - 40 = 0
t^2 + 8t - 5t - 40 = 0
t(t+8) - 5(t+8) = 0
t = 5 & t = -8
t = 5 accepted. (Ans)

ii) v = u + at
v = o + 0.5(5)
v = 2.5 (Ans)

Thanks.

 

[Maayee]

Can anybody help me with this question?
A car is travelling along a straight road. It accelerates uniformly from rest to a speed of 15ms^-1 and maintains this speed for 10 minutes. It then decelerates uniformly to rest. If the acceleration and deceleration are 5ms^-2 and 8ms^-2 respectively, find a total journey time and total distance travelled during the journey.

 

[The Sarcastic Retard]

Can anybody help me with this question?
A car is travelling along a straight road. It accelerates uniformly from rest to a speed of 15ms^-1 and maintains this speed for 10 minutes. It then decelerates uniformly to rest. If the acceleration and deceleration are 5ms^-2 and 8ms^-2 respectively, find a total journey time and total distance travelled during the journey.

What is answer?

 

[Maayee]

i already got it, the answer for the time was 605s and the distance 9075m

 

[The Sarcastic Retard]

i already got it, the answer for the time was 605s and the distance 9075m

Post the solution.

 

[Maayee]

When the car is accelerating:

s = (v²-u²)/2a
= (15²-0)/2(5) = 22.5 m
time taken for acceleration
t = (v-u)/a
= (15-0)/5 = 3 s

When the car is speeding uniformly:
s = velocity x time = 15 x 10 min x 60s = 9000m

When the car is decelertating:

s = (v²-u²)/2a
= (0-15²)/2(-8) = 14.1 m
time taken for decelartion
t = (v-u)/a
= (0-15)/(-8) = 1.9 s


Total journey time = 3 + (10x60) + 1.9 = 604.9s

 

[Maayee]

then distance is speed x time so 604.9 x 15 = 9073.5m