Maths, Addmaths and Statistics: Post your doubts here!

[ZohaKhan]

ok il see
Bi is still nt clear by d way

 

[Asad.Babur]

View attachment 38071

i dunno i just typed the answer


View attachment 38071
i) just find the area of this shaded cylindrical region
ii) 1 cubic cm = 1 ml
and 25000 l = 25000000 ml

so time taken for 1 ml (cubic cm) to come out = 1 sec
thus time taken for 25000000 ml = 25000000/(Volume u found in 'i')

you are Wrong here! I presume

 

[Asad.Babur]

ok il see
Bi is still nt clear by d way

Wait!!!

 

[Asad.Babur]

ok il see
Bi is still nt clear by d way

(i) Formulae here is = Pi*Speed*Square of Radius

So, in this case
Pi =Moslty 3.142, but Varies Calculator to Calculator
Speed = 300cm/s as given
Sqaure of Radius = 4.5 * 4.5 = 20.25

hence we have, 3.142 (Pi) x 300 (Speed) x 20.25 (Sqaure of Radius)= 19087.65cm/s

Lets Simplyfy =19088 cm/s or 19.1 liters

Whats Confusing?

 

[Asad.Babur]

you are Wrong here! I presume

No, You are right in Seconds, just convert to mins

 

[Asad.Babur]

ok il see
Bi is still nt clear by d way

I said Pi * Speed * Raduis^2

3.142 * 300 * 4.5 ^2=

3.142 * 300 * 20.25 = 19100 Cm^3

 

[Awesome12]

View attachment 38071

i dunno i just typed the answer


View attachment 38071
i) just find the area of this shaded cylindrical region
ii) 1 cubic cm = 1 ml
and 25000 l = 25000000 ml

so time taken for 1 ml (cubic cm) to come out = 1 sec
thus time taken for 25000000 ml = 25000000/(Volume u found in 'i')

Paen kera software hai

 

[Asad.Babur]

I never understand, what is so confusing in this? ZohaKhan

 

[***amd***]

Paen kera software hai

software?

 

[ZohaKhan]

I never understand, what is so confusing in this? ZohaKhan

got u Sorry sorry

 

[***amd***]

got u Sorry sorry

u really got that?

 

[ZohaKhan]

Yes

 

[Asad.Babur]

got u Sorry sorry

Oh! Finally, Useless me came Usefull

 

[Dream.Eater]

Can someone tell me how to do Q25

 

[***amd***]

Can someone tell me how to do Q25

as we know that

the area under the speed-time graph is the distance traveled by the object.

(a)

so in first 20 seconds the car travels D meters. that means, 0.5 * u * 20 = D (by applying the formula for area of a triangle.
10u = D

for the time after '20 seconds' we need to find when does the car travels another D meters.
so t * u = D
t * u = 10u
so we divide both sides of equation by u and we get, t = 10
thus 2D meters are traveled in 20 + 10 = 30 seconds
(b)
acceleration from 0 to 20 seconds = (u-0)/(20-0) = u/20
thus if we change acceleration to deceleration (negative acceleration: in which the speed slows down) we just change its sign i.e.
(-u/20) * 0.5 (coz Q says deceleration was half of it) = -u/40
so according to the formula acc = v/t
we get t = v/acc by dividing both sides by acc and then multiplying them with t.
so t = 60 + [(u/4) - u] / -u/40
where [u/4] - u = velocity change from 60 seconds to t seconds

hope you got it

 

[Gohar Awais]

For two similar triangles like these how do I determine the unknown sides? Tha angle at B is a right angle and so is ADB a right angle

AD= q cm
AB=p cm
AL= q+9 cm
BL=15 cm
BD=12 cm
DL=9 cm

please reply fast
I can't determine which sides are corresponding to each other

 

[***amd***]

For two similar triangles like these how do I determine the unknown sides? Tha angle at B is a right angle and so is ADB a right angle

AD= q cm
AB=p cm
AL= q+9 cm
BL=15 cm
BD=12 cm
DL=9 cm

please reply fast

View attachment 38213

for any pair of triangles, u should take:
base : base
hypotenuse : hypotenuse
height : height
all i mean to say is, u should take ratios of corresponding sides.

as hyp (hypotenuse) is always the side opposite to right angle
so we take the ratio p : (9 + q)

the smaller side of each triangle...
12 : 15

longer side making the right angle...
q : p

we deduce
12/15 = p/(9+q) ----------> 1

12/15 = q/p -----------> 2


solve the 2 simultaneous equations and u'll get ur answer

 

[Gohar Awais]

thanks man

 

[Asad.Babur]

as we know that
(a)
View attachment 38210
so in first 20 seconds the car travels D meters. that means, 0.5 * u * 20 = D (by applying the formula for area of a triangle.
10u = D

for the time after '20 seconds' we need to find when does the car travels another D meters.
so t * u = D
t * u = 10u
so we divide both sides of equation by u and we get, t = 10
thus 2D meters are traveled in 20 + 10 = 30 seconds
(b)
acceleration from 0 to 20 seconds = (u-0)/(20-0) = u/20
thus if we change acceleration to deceleration (negative acceleration: in which the speed slows down) we just change its sign i.e.
(-u/20) * 0.5 (coz Q says deceleration was half of it) = -u/40
so according to the formula acc = v/t
we get t = v/acc by dividing both sides by acc and then multiplying them with t.
so t = 60 + [(u/4) - u] / -u/40
where [u/4] - u = velocity change from 60 seconds to t seconds

hope you got it

lolX! Man I got it atleast

 

[Dream.Eater]

as we know that
(a)
View attachment 38210
so in first 20 seconds the car travels D meters. that means, 0.5 * u * 20 = D (by applying the formula for area of a triangle.
10u = D

for the time after '20 seconds' we need to find when does the car travels another D meters.
so t * u = D
t * u = 10u
so we divide both sides of equation by u and we get, t = 10
thus 2D meters are traveled in 20 + 10 = 30 seconds
(b)
acceleration from 0 to 20 seconds = (u-0)/(20-0) = u/20
thus if we change acceleration to deceleration (negative acceleration: in which the speed slows down) we just change its sign i.e.
(-u/20) * 0.5 (coz Q says deceleration was half of it) = -u/40
so according to the formula acc = v/t
we get t = v/acc by dividing both sides by acc and then multiplying them with t.
so t = 60 + [(u/4) - u] / -u/40
where [u/4] - u = velocity change from 60 seconds to t seconds

hope you got it

Thanks!