doubleDx:


From LHS

500 => 20(14) + 22(10)

Since LHS = RHS
Comparing LHS & RHS yields =>

20(14) + 22(10) = 20x + 22y
14 + 10 = x + y
thus =>
x = 14 and y = 10

bolkay47:
yes sir.. But I used lagragian multiplier and I got different answer and it satisfies the equation like yours.

doubleDx:


Yes, that's what I was thinking too! There many answers that satisfies the equation....so, whichever works for you is okay!

bolkay47:
Try this :::: 500=20x+22y. Find x and y

DrMaths:
Hello Guys,
Forget about mah moniker

Plz help me with dis
Z = ¡^1001

Av tried all mathematical tricks i kw, but i stil got stooked!

agentofchange1:
oo my bad, it just dawn on me , there was a mistake in the substitution

i will post the correction later . chai ..

miniyi:

Will be expecting.. Thanks

Laplacian:
Find the differential equation corresponding to;
y=Ae^3x+Be^x.............(1)

Solution:
since there are two unknown constants (A & B), we need two differential equations.
y'=3Ae^3x+Be^x..............(2.)

y"=9Ae^3x+Be^x................(3.)

if, from eqn(1), we substitute for Be^x in (2) and (3) we get;
y'=2Ae^3x+y...................(4)

y"=8Ae^3x+y...................(5)

now, y"-4*eqn(4) gives
y"-4y'+3y=0

Abdullahi4u7:
Pls explain with an example the concept of arithmetic and geometric progressions.

calculus3e:
integrate sinXcosX

calculus3e:
find the value of 5e^0.5 to 5significant figure

rhydex247:

$sinxcosx dx
using substitution method
let u=cosx.... du=-sinxdx..... dx= -du/sinx
$ sinx(u) -du/sinx
-$udu.... -u^2/2 + C
finally, we have -cos^2 x/2 + C

benji93:

you tried bro,but maybe you should try it this way
sin(2x) =2sinxcosx
(1/2)sin(2x) =sinxcosx
dy = sinxcosx dx= (1/2)sin(2x) dx
y=(-1/24)cos(2x)

jackpot:
@Maths generals,

Please help meeeeeee


A combat-ready soldier on a level ground, wants to toss a grenade from a height of 1.8 meters off the ground, to a stationary enemy target, on the edge of a cliff of height 50metres. The horizontal distance of the soldier from the foot of the cliff is 20metres.

(a) If he intends tossing at an angle of 75degrees, what is the least initial velocity of his projection?

(b)If he intends tossing with an initial velocity of 70m/s, what is the least angle of projection if his missile will hit the target at the shortest possible time?

(take g=9.8m/s², ignore spin, drag force, air resistance)



Tags: agentofchange1, doubleDx, Laplacian, AlphaMaximus, STENON, AmazingAngel, Richiez, rhydex247, montty, CalculusFx, thankyouJesus, calculus3e, all maths generals, sergeants

agentofchange1:


OK , on bed now without writing stuffs , maybe 2maro b4 nw..

jackpot:


A combat-ready soldier on a level ground, wants to toss a grenade from a height of 1.8 meters off the ground, to a stationary enemy target, on the edge of a cliff of height 50metres. The horizontal distance of the soldier from the foot of the cliff is 20metres.

(a) If he intends tossing at an angle of 75degrees, what is the least initial velocity of his projection?

(take g=9.8m/s², ignore spin, drag force, air resistance)

tohero:

Using the formula v² = u²+2gh
Where
v=0 ,
g=-9.8m/s²
h= 50 - 1.8 = 48.2m
Vertical component of the initial velocity = u sinA
Where A=75^o.
Then the formula bcoms
0² = u²sin²75^o - 2*9.8*48.2
That is
19.6*48.2 = u²sin²75^o
Calculator should finish this.

agentofchange1:


let's differentiate his answer & see if we could get back the integrand

d(-cos^2 x/2) /dx = -(-2) cosxsinx /2 = cosxsinx = sinxcosx..............(as expected)



now let's check yours

dy/dx = -(-2)/24 sin2x = 1/12 sin2x = 2/12 * sinxcosx

=1/6 sinxcosx =/= sinxcosx ??.

so what can you say now ?.

jackpot:
but Sir, does it mean that the distance (20m) of the cliff from the soldier doesn't matter in the solution?

rhydex247:

$sinxcosx dx
using substitution method
let u=cosx.... du=-sinxdx..... dx= -du/sinx
$ sinx(u) -du/sinx
-$udu.... -u^2/2 + C
finally, we have -cos^2 x/2 + C

jackpot:
Question four



A poacher standing at a distance of 25m from a tree saw, aims and takes a shot with a catapult at a bird whose vertical distance from the ground is 30m. Suppose that
(i) the catapult is held at a distance of 1.5m above ground level,
(ii) the initial velocity of the shot was 40m/s,
(iii) the bird was hit before it could make a move,
then calculate
(a) the duration of the shot
(b) the angle of the shot