I really need help with this question.

Two circles touch externally at T. A chord of the first circle XY is produced and touches the other at Z. The chord ZT of the second circle, when produced, cuts the first circle at W. Prove that angle XTW = angle YTZ.

Thanks in advance.

HAY guys.....greet thee all

its BENBUKS ....



THIS's NOW MA NEW ACCOUNT ...

NICE SOLVING..


-SHALOM-...

jackpot:
ooh, my benbuks. Where art thou?

hay "miss" jackport , honey i'm here . anything for the boys . ?

Calculusfx:
...smile,master rhydex,i wish i had ur time,so busy these days,all the way...my greetings to all the generals on the thread,JACKPOT,LAPLACIAN,BENBUKS,RHYDEX,FACTORIAL and others.

hay prof.

its "benbuks" greet thee and other generals..


1luv..

wow well done guys
don't forget to like our page where you will learn more about science and technology, also be our top scholar for this week. centre for future leader.
https://www.facebook.com/pages/Centre-for-Future-Leaders/260662524082509

This is d Lord's doing and it is marvelous in my sight....People Are Goooooooodddddd.I throway salute oo

hello guys...which of this courses would you advice someone to study..pure and applied maths or statistics..pls state some reasons..thanks

pls help with this...
Out of 500 families with 5 children each, how many would you expect to have:
(a)exactly 2 boys
(b)more than 1 boy
(c)less than 2 boys
roundup ur answer tothe nearest whole number.

sundaylinus:
pls help with this...
Out of 500 families with 5 children each, how many would you expect to have:
(a)exactly 2 boys
(b)more than 1 boy
(c)less than 2 boys
roundup ur answer tothe nearest whole number.

Lets use

E(x) = m =Np

known that 1= q+p

P(boy) =P(girl) =1/2

1/2 =p =q

N=500

n=5

by binomial distribution

P(X=r) = nCr*q^(n-r) * p^r

a) P(X=r=2) =5C2*(1/2)⁵ = 10/32

we would expect (500)(10/32) =5000/32 … finish up

b) P(r>1) =p(2) +p(3)+p(4) +p(5) = .... finish up then multiply your answer by 500

c) p(r <2) = p(0) +p(1) ..... similar as former




hope you get...

benny1620:


Lets use

E(x) = m =Np

known that 1= q+p

P(boy) =P(girl) =1/2

1/2 =p =q

N=500

n=5

by binomial distribution

P(X=r) = nCr*q^(n-r) * p^r

a) P(X=r=2) =5C2*(1/2)⁵ = 10/32

we would expect (500)(10/32) =5000/32 … finish up

b) P(r>1) =p(2) +p(3)+p(4) +p(5) = .... finish up then multiply your answer by 500

c) p(r <2) = p(0) +p(1) ..... similar as former




hope you get...

pls. Benny put more light, this topic was introduced to us by this assignment and we are to submit it on 20th pls.

sundaylinus:

pls. Benny put more light, this topic was introduced to us by this assignment and we are to submit it on 20th pls.

ok bro , didn't have much time , was in the lecture hall .
But go get a probability /statistics text book for details .

Now on the question , considering 500 families ( maybe in Lagos ) we're told that each of them have 5 children . could be all boys, or all girls or mixed .Using probability techniques , we want to predict / say on average the number of boys the family could consist of .

we use binomial distribution ,

p= probability of a boy in a family

q= probability of NOT a boy ( i.e a girl)

since we know that most usually , you're either a boy or a girl ( ignore special cases ) , just like tossing a coin , we either have 'heads' or 'tails' ( i.e only two possible outcomes ) , thus in that case , we have equal probabilities

we should know that p+q=1 ...........( from axioms)

p=1/2 ........(probability of success )

q= 1 -p = 1- 1/2= 1/2 ....(probability of failure )

now , we see that

p=q=1/2

from here go to my initial solution above .

follow it carefully and finish up , you don't expect me to do everything for you do you.? or you tell me where you're confused .


gracia...

Good afternoon Gurus in the house . please help with this trigonometry and analytical geometry questions .
1. cosx + cos3x+cos5x+cos7x=0
2. cos2x+cosx+1=0
3.sin30+sin2tita =0

Thanks in advance

Good day to u all, it has been a while since I visited this great thread, I am the former 'boladearo' a student of all the maths generals on this thread.

Find the quadratic equation which has the difference of its roots to be 2, and difference of the square of its roots to be 5.
Soln.
X - y = 2 ...............1
X^2 - y ^2 = 5 .....2
X = 2 + y .............3
Subt d value of x into eqtn 2
(2+y)^2 - y^2 = 5
4 + 4y + y^2 - y^2 = 5
4y = 5 - 4
Y = 1/4 . X =9/4
X+y =5/2. xy =9/16
D eqtn is then x^2 - 5/2x + 9\16
16x^2 - 40x + 9
Soln 2
X-y = 2
X^2 - y^2 = 5
(X-y)^2 + 2xy = 5
2^2 + 2xy = 5
2xy = 5 - 4
Xy = 1/2
(X-y)(x+y) = 5
2(x+y) = 5
X + y = 5/2
So equtn is x^2 - 5/2x + 1/2 =2x^2 -5x +1

Pls ma generals in the house why is ma ans not the same. I ought to get the same ans since I followed maths rule in all ma solving .

edaloba:
Find the quadratic equation which has the difference of its roots to be 2, and difference of the square of its roots to be 5.
Soln.
X - y = 2 ...............1
X^2 - y ^2 = 5 .....2
X = 2 + y .............3
Subt d value of x into eqtn 2
(2+y)^2 - y^2 = 5
4 + 4y + y^2 - y^2 = 5
4y = 5 - 4
Y = 1/4 . X =9/4
X+y =5/2. xy =9/16
D eqtn is then x^2 - 5/2x + 9\16
16x^2 - 40x + 9
Soln 2
X-y = 2
X^2 - y^2 = 5
(X-y)^2 + 2xy = 5
2^2 + 2xy = 5
2xy = 5 - 4
Xy = 1/2
(X-y)(x+y) = 5
2(x+y) = 5
X + y = 5/2
So equtn is x^2 - 5/2x + 1/2 =2x^2 -5x +1

Pls ma generals in the house why is ma ans not the same. I ought to get the same ans since I followed maths rule in all ma solving .

the bold expression is where you got the second solution wrong. x^2 - y^2 =/= (x - y) ^2 + 2xy(not equal)
(x - y) ^2 + 2xy = x^2 + y^2

mathefaro:
the bold expression is where you got the second solution wrong. x^2 - y^2 =/= (x - y) ^2 + 2xy(not equal)
(x - y) ^2 + 2xy = x^2 + y^2

thank you very much sir.
Hav seen ma mistake

There's a new smartphone app for solving maths. you can download it here http://www.figureout.com.ng/photomath-download-phone-app-can-solve-mathematics/

pls the gurus in the house I need yur assistance on this assigment and please I will appreciate it if the solutions are accompanied with explanation :
1.determine the radius and center coordinate of the circle whose equation is 4x^2+4y^2-16x+32y-116=0

2.find the gradient of the tangent to the curve. 3x^2.y^3-5x^3.y+8x=11y^2-3x^3. at(1,1)and (0,-2).

3. evaluate
lim. (e^-×_1/x)
x--0

4.given f(x)={x-5. X<a
.......................{7/x+1. x>=a
find 'a' such that f(x) is continuous at. x=a

5.evaluate . (a) ∫e^-x sin3xdx.
(b)∫^2. x^-2in(2x)dx
....5 (pls note that the 5 is an underscore on the integration sign)
now over to my bosses.benbuks,jackpot,rhydex,calculus
fx,efficiency,laplacian........

5.evaluate . (a) ∫e^-x sin3xdx.

solution.
Let I=∫e^-x sin3xdx.
using integration by part.
let u=sin3x and dv=e^-x
du=3cos3xdx and v=-e^-x
UV-$VDU
{-e^-x}{sin3x} -$ {-e^-x}{3cos3x}dx.
-e^-xsin3x + 3${e^-x}{cos3x}dx
similarly, let u=cos3x and dv=e^-x
du=-3sin3xdx and v= -e^-x
-e^-xsin3x + 3{ uv-$vdu}
-e^-xsin3x + 3{(cos3x)( -e^-x) -$ (-e^-x)(-3sin3x)dx}
-e^-xsin3x + 3{-e^-xcos3x- 3$(e^-x)(sin3x)dx}
recall dat I=∫e^-x sin3xdx.
I=-e^-xsin3x + 3{-e^-xcos3x- 3(I)}
I=-e^-xsin3x- 3e^-xcos3x-9I
10I=-e^-xsin3x- 3e^-xcos3x
I=1/10 (-e^-xsin3x- 3e^-xcos3x)
I=-1/10(e^-xsin3x+3e^-xcos3x)

@rhydex, thanks for that I will appreciate it if u'll assist me in solving the rest of the questions. still calling on other gurus in the house to cum to my rescue @..bosses.benbuks,jackpot,calculus
fx,efficiency,laplacian.

1.determine the radius and center
coordinate of the circle whose
equation is 4x^2+4y^2-16x
+32y-116=0
Solutn
Recall dat d eqn of a circle is given by
x^2+y^2+2gx+2fy+c=0........ eqn *
4x^2+4y^2-16x+32y-116=0 (dividing thru by 4) yield
x^2+y^2-4x+8y-29=0......... eqn **
by comparing eqn * &** we av
g=-2
f=4
c=-29
Now, for centre = (-g,-f)=(2,-4).
also radius= √(g^2+f^2-c) =√(-2)^2 +4^2+29= √49=7
clearly, centre= (2,-4) and radius=7.

3. evaluate
lim. (e^-×_1/x)
x--0

(b)∫^2. x^-2in(2x)dx
....5 (pls note that the 5 is an
underscore on the integration sign)

2.find the gradient of the tangent to
the curve. 3x^2.y^3-5x^3.y
+8x=11y^2-3x^3. at(1,1)and (0,-2).
please i will really appreciate if u can re- write these three questions in an appropriate way.

Soneh:
pls the gurus in the house I need yur assistance on this assigment and please I will appreciate it if the solutions are accompanied with explanation :
1.determine the radius and center coordinate of the circle whose equation is 4x^2+4y^2-16x+32y-116=0

2.find the gradient of the tangent to the curve. 3x^2.y^3-5x^3.y+8x=11y^2-3x^3. at(1,1)and (0,-2).

3. evaluate
lim. (e^-×_1/x)
x--0

4.given f(x)={x-5. X<a
.......................{7/x+1. x>=a
find 'a' such that f(x) is continuous at. x=a

5.evaluate . (a) ∫e^-x sin3xdx.
(b)∫^2. x^-2in(2x)dx
....5 (pls note that the 5 is an underscore on the integration sign)
now over to my bosses.benbuks,jackpot,rhydex,calculus
fx,efficiency,laplacian........

#2. Solving by implicit differentiation

>>> 3(x^2 . 3y^2 dy/dx + y^3 . 2x) - 5(x^3 dy/dx + 3x^2 . y) + 8 = 22y dy/dx - 9x^2

Opening the Bracket
>>>> 9x^2 . y^2 dy/dx + 6xy^3 - 5x^3dy/dx - 15x^2 . y + 8 = 22y dy/dx - 9x^2

Collect Like terms,then factorize dy/dx out
>>>> dy/dx (9x^2 . y^2 - 5x^3 - 22y) = 15x^2 . y - 6xy^3 -8 - 9x^2

>>>> dy/dx = (15yx^2 - 6xy^3 - 8 - 9x^2)/(9x^2 . y^2 - 5x^3 - 22y)

>>>> For the first cordinate (1,1) ,Substituting x = 1 and y =1 into the above equation
dy/dx = 2/9.
>>>> For the 2nd coordinate (0, -2), substituting x = 0 and y = -2 into the same equation
dy/dx = -2/11.

rhydex247:
1.determine the radius and center
coordinate of the circle whose
equation is 4x^2+4y^2-16x
+32y-116=0
Solutn
Recall dat d eqn of a circle is given by
x^2+y^2+2gx+2fy+c=0........ eqn *
4x^2+4y^2-16x+32y-116=0 (dividing thru by 4) yield
x^2+y^2-4x+8y-29=0......... eqn **
by comparing eqn * &** we av
g=-2
f=4
c=-29
Now, for centre = (-g,-f)=(2,-4).
also radius= √(g^2+f^2-c) =√(-2)^2 +4^2+29= √49=7
clearly, centre= (2,-4) and radius=7.

Another way, so u ve a choice
The equation of a circle cud be written as
(x - a)^2 + (y -b)^2 = r^2
The idea is trying to reduce the equation u were given to look like the general equation of a circle,so u cud relate them,then find the centre and radius.

4x^2 + 4y^2 - 16x + 32y = 116
Divide through by 4
x^2 + y^2 - 4x + 8y = 29

x^2 - 4x + y^2 + 8y = 29

Using completing the square method to simplify
(x^2 -4x + 4) + (y^2 + 8y + 16) = 29 +4 + 16

(x - 2)^2 + (y + 4)^2 = 49

Relating the above equation to the equation of a circle
a = 2
b = -4
r^2 = 49
r = √49
r = 7
Centre (a,b) = (2 , -4)
Radius (r) = 7.

Please help!!
Angle of elevation:
-an engineer wishes to bridge a river.he observe that the angle of elevation of the top of a tree on the far bank is 27degree from a point on the near bank and 14degree from a point 30m further back.. How wide was the river Thanks
(2)from point P on level ground the angle of elevation of the top of a tower is 26degree 51'(min),from a point 25m closer to the tower and on the same line with p and the base of the tower of the angle of elevation of the top is 53degree 30min.... Calculate to three significant figures the height of the tower..
Please help me drawing(photo) will be highly appreciated. Thanks

rhydex247:
3. evaluate
lim. (e^-×_1/x)
x--0
.
t as x tends to zero of e. to the power -x minus 1 all divided by x
.
(b)∫^2. x^-2in(2x)dx
....5 (pls note that the 5 is an
underscore on the integration sign)
Integration sign with 2 as superscore and five as underscore of x to the power -2In(2x)dx

2.find the gradient of the tangent to
the curve
. 3X²y³-5X³Y+8x=11Y² -3X³ at (1,1)and (0,-2).
please i will really appreciate if u can re- write these three questions in an appropriate way.

Sorry for replying late, hope the modifications gives you a clearer understanding of the questions

FuckYou:
Please help!!
Angle of elevation:
-an engineer wishes to bridge a river.he observe that the angle of elevation of the top of a tree on the far bank is 27degree from a point on the near bank and 14degree from a point 30m further back.. How wide was the river Thanks
(2)from point P on level ground the angle of elevation of the top of a tower is 26degree 51'(min),from a point 25m closer to the tower and on the same line with p and the base of the tower of the angle of elevation of the top is 53degree 30min.... Calculate to three significant figures the height of the tower..
Please help me drawing(photo) will be highly appreciated. Thanks

Hello, Mr FuckYou!

Soneh:
pls the gurus in the house I need yur assistance on this assigment and please I will appreciate it if the solutions are accompanied with explanation :
1.determine the radius and center coordinate of the circle whose equation is 4x^2+4y^2-16x+32y-116=0

baby question. Already answered.

2.find the gradient of the tangent to the curve. 3x^2.y^3-5x^3.y+8x=11y^2-3x^3. at(1,1)and (0,-2).

wrongly posed question since the points (1, 1) and (0, -2) doesnt even lie on the given curve.

3. evaluate
lim. (e^-×_1/x)
x--0

not clear.

4.given f(x)={x-5. X<a
.......................{7/x+1. x>=a
find 'a' such that f(x) is continuous at. x=a

solve x-5= 7/x + 1 to get the value(s) of a.

5.evaluate . (a) ∫e^-x sin3xdx.
(b)∫^2. x^-2in(2x)dx
....5 (pls note that the 5 is an underscore on the integration sign)
now over to my bosses.benbuks,jackpot,rhydex,calculus
fx,efficiency,laplacian........

a is baby.

For b, express x^-2 ln(2x) as (ln 2)/(x^2) + (ln x)/(x^2) and go from there.

Pls help me out on this, showing out clear workings

Pls help me out on this, showing out clear workings
(1)Calculate the mid point of the line of segment. Y - 4X + 3 = 0 which lies between the X-axis and Y-axis


(2)Find the equation of the straight line through (-2,3) and perpendicular to 4X + 3Y - 5 = 0

jackpot:

baby question. Already answered.

wrongly posed question since the points (1, 1) and (0, -2) doesnt even lie on the given curve.

not clear.

solve x-5= 7/x + 1 to get the value(s) of a.
a is baby.

For b, express x^-2 ln(2x) as (ln 2)/(x^2) + (ln x)/(x^2) and go from there.

this girl... I just don't know wat to say

tuwayz:
Pls help me out on this, showing out clear workings
(1)Calculate the mid point of the line of segment. Y - 4X + 3 = 0 which lies between the X-axis and Y-axis


(2)Find the equation of the straight line through (-2,3) and perpendicular to 4X + 3Y - 5 = 0

1. Given: Y-4x+3=0

when y=0,
0-4x+3=0
-4x=-3
=>>> x=3/4
the point is thus =(3/4,0)

when x=0,
y-4(0)+3=0
y =-3
=>>> y=-3
thus the point is =(0, -3)

but, midpoint=(x1+x2)/2, (y1+y2)/2

for x, (3/4+0)/2=3/8
for y , (-3+0)/2 =-3/2

>>> therefore, midpoint is (x,y) = (3/8,-3/2)

2. Given 4X + 3Y - 5 = 0

we proceed by putting it in gradient intercept form
i.e , Y= -4/3X +5/3

for 2 lines to be perpendicular, m1 * m2 = -1

thus, -4/3 * 3/4 will = -1

so new gradient for perpendicular will be 3/4

from formula for gradient,

(y-y1)/(x-x1) = m2

so, 4(y-3) = 3(x+2)

4y -12= 3x+6
=>>>>>> 4y=3x+18