- Messages
- 455
- Reaction score
- 7,676
- Points
- 503
For the first time I saw this. :3
Anyways ty. Though I am not getting why we did like this.
Can u help me wid other doubts as well?
Thanks.
& Sure. Inbox them or post them here
Last edited:
We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
For the first time I saw this. :3
Anyways ty. Though I am not getting why we did like this.
Can u help me wid other doubts as well?
Thanks.
Thanks for the help. I have posted here, check my post.& Sure. Inbox them or post them here
Q11. bThanks for the help. I have posted here, check my post.
Answer is no solution.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
Thats because absolute value is always positive So just write a brief statement along with your answerAnswer is no solution.
Q15 aThanks for the help. I have posted here, check my post.
How? Its correcct.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.
Make the RHS equal to zero.How? Its correcct.
Thanks.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
Use standard normal distributionHow to solve this normal distribution?
P(|Z|>2.4)
The answer us 0.0164
Please tell me which identities to use here...
Let distances be s1 and s2 (use s = ut + 1/2at^2)
Thanks.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)
What is answer?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.
Post the solution.i already got it, the answer for the time was 605s and the distance 9075m
For almost 10 years, the site XtremePapers has been trying very hard to serve its users.
However, we are now struggling to cover its operational costs due to unforeseen circumstances. If we helped you in any way, kindly contribute and be the part of this effort. No act of kindness, no matter how small, is ever wasted.
Click here to Donate Now