Dibu_Ojerinde:

F = m x a,
F = ^{m x v}/_t
V = 20ms^-1
so we have to find mass and time
density = 1mgm^-3
area = 0.0005m²
from the denstity, if mass = 1mg, volume = 1m³
so length = ^volume/_area
= 2000m
if length = 2000m,
20 x time = 2000
time = 100s
Mass = 1mg = 1 x 10⁶kg

F = (1 x 10^-6) x 20 / ₁₀₀
= 0.2 x 10^-6N

....please correct me sir

Mr Calculus: RECHECK UR QUES@LAZ

Dr. Laz'Devitan:


I have problem with your question for some reasons
1)waec don't set problems of equation great than power degree 2 in
ordinary pure mathematics.So,i guess...it may be a further
mathematics(as u guys may put it)
2) I got problem with the smiley's in ur question too!

However,this may set-out to be way for the solution!...after the
expansion i got

x^3 + x^2 - 2x - 8=Mx + Mc.......

The next thing would be to work by coefficient of x....which we can accept as

-2x=Mx.....
thus,M=-2

taiocol: WAEC 2013:
(x-2)(x²+3x+2)+2(x+2)(x-1)=(x+2)M
FIND M

Dr. Laz'Devitan:


Very wrong!

Dr. Laz'Devitan:


Using the equation provided by....from the equation...we can break x^2 + 3x + 2 into/equal to (x+1)(x+2)....so it bcoms

(x-2)(x+1)(x+2) + 2(x+2)(x-1)=M(x+2).....Looking the equation..u will observe that x+2 is common..so it cancels out...it bcoms
(x-2)(x+1) + 2(x-1)=M
x^2 + x -2x - 2 + 2x - 2 = M
x^2 + x - 4 = M
x^2 + x -( 4 + M)=0.......If we were to solve for the root using trial error rule......we shall find to our surprise that

M=2!

taiocol: i got same ansa nd frnd said i failed. He refused 2 solve it.

taiocol: i got same ansa
nd frnd said i failed. He refused 2 solve it.

Dr. Laz'Devitan:
........This question goes out to engineering student mostly!

Find the values of y given that :
(y^2 - 9y + 15)(y^2 - 9y + 20)=6

aysuccess99:
solution
This question would need the application of polynomial.
We would first, expand the two equations i.e, (y^2-9y+15)(y^2-9y+20)=6
when we expand we have:
y^4-9y^3+20y^2-9y^3+81y^2-180y+15y^2-135y+300=6
then we collect like terms, we now have:
y^4-18y^3+116y^2-315y+300=6
we would now conclude our equation by bringing 6 to the RHS, we have.
y^4-18y^3+116y^2-315y+300-6=0
we then have:
y^4-18y^3+116y^2-315y+294=0
this is where the application of polynomial sets in.
so the first value that can divide and give us 0 in the equation is 2.
f(x)=y^4-18y^3+116y^2-315y+294.
when x=2, we have 0 as remainder.
f(2)=(2)^4-18(2)^3+116(2)^2-315(2)+294
f(2)=16-18(+116(4)-630+294
f(2)=16-144+464-630+294
f(2)=0
therefore, the first value of y=2

aysuccess99: continuation
y=2 is the same as y-2=0
we then divide f(x) by (y-2), i.e y^4-18y^3+116y^2-315y+294 by (y-2)
so we would have y^3-16y^2+84y-147 as our new f(x)
we would then find the value of x that can divide our new f(x) to give us another value.
after trying all values ranging from 1 to 6, we discover that the value that can make f(x)=0 is 7.
f(x)=y^3-16y^2+84y-147
f(7)=(7)^3-16(7)^2+84(7)-147
f(7)=343-16(49)+588-147
f(7)=343-784+588-147
f(7)=0, so the next value that would give us 0 is 7
therefore, the new value that would give us 0 is 7
therefore, y=7.

aysuccess99: the remaining equation is y^2-9y+21, but it is impossible to get the value of y here.
so all the values of y is now 2,7 and y^2-9y+21
i hope i am correct dr. lezdevitan

Ai-bi-ke:
H CL
| |
H-C-C ---C=O
| | |
H H H-C-H
|
H

what is d IUPAC name?

aysuccess99: the remaining equation is y^2-9y+21, but it is impossible to get the value of y here.
so all the values of y is now 2,7 and y^2-9y+21
i hope i am correct dr. lezdevitan

Dr. Laz'Devitan:
........This question goes out to engineering student mostly!

Find the values of y given that :
(y^2 - 9y + 15)(y^2 - 9y + 20)=6

Dr. Laz'Devitan:

Let y^2 - 9y=M....then,
(M+15)(M+20)-6=0
M^2 + 35M + 294=0
(M+21)(M+14)=0......
so,y^2 - 9y=-14
Finally..u land at the answer!

Kunlexic: are u d one called ''iwe'' coz diz ur way of solving phy,maths and chem questions fear me.so wat course are u aspiring 4 and wic schl?...wish u success.

Dr. Laz'Devitan:


The name of the compound is

3-CHLOROPROPAN-2-ONE!

Dr. Laz'Devitan:
......Let me again pinch u all with a good chemistry question

Balance the following REDOX REACTIONS

1)[CLO(3)]^(-1) + Zn ---> Cl^(-1) + Zn^(+2)

Dr. Laz'Devitan:
......Let me again pinch u all with a good chemistry question

Balance the following REDOX REACTIONS

1)[CLO(3)]^(-1) + Zn ---> Cl^(-1) + Zn^(+2)

great.david:
Find the acceleration after 3 seconds of a body moving in a straight line with this equation y=4t^4 - 3t^3

aysuccess99:
this looks like a further maths question, but I will do answer it.
solution
x=4t^4-3t^3
V=dx/dt=16t^3-9t^2
therefore, V is velocity, and it is 16t^3-9t^2
and acceleration=velocity/time
a=dv/dt
a=48t^2-18t
therefore, when t=3seconds, acceleration would be.
a=48(3)^2-18(3)
a=48(9)-18(3)
a=432-54
a=378m/s^2
therefore,:::::::: the acceleration after 3seconds is 378m/s^2
to the bosses and ogas in the thread, my answer is subjected to correction. Thanks.