Richiez:

oh i'm very very sorry i got the wrong impression....okay we all are here to help you okay?

Ghettoguru:

Sorry buddy, we initially got the wrong impression. Keep your questions coming, there are a lot of gurus on this thread willing to help!

johnpaul1101: ok richiez, i'm reliefed now, but i will try 2 reduce d question, but i hav one problem;logical reasoning.....wen it comes 2 logical questions such as WORD PROBLEM LEADING TO........it wud b like my brain is locked, i.e.....d incapability of representing an equation from a word question(written question)

Chelcy10: Make x the subject of the formula; STX^3-2WX=10X^2T

Chelcy10: Make x the subject of the formula; STX^3-2WX=10X^2T

Chelcy10: Make x the subject of the formula; STX^3-2WX=10X^2T

Johncasey: Evaluate d cube root of (10+6*sqrt of 3)

Richiez:

Question 1
Now we have to rewrite the straight line eqn
x + y = 7
x + y - 7 = 0 ................(1)
For any straight line Ax+By+c=0 and a point (m,n) on same plane, the perpendicular distance between the point and the plane is given by;
d = (lAm+Bn+Cl)/√(A^2 + B^2)
obviously, A=1 , B=1 , C+ -7 , m=4 , n=5
therefore;
d = (l4+5-7l)/√(1^2 +1^2)

d = l2l/√(2) = 1.414units(approx)

Question 2
here, the main task is to find the coordinates of the point on line AB, where the line L cuts it in the ratio 2:3

Let the coordinate of the point be P(x,y)
For the line AB;
A(x1,y1) = A(-1,4)
B(x2,y2) = B(4,-1)

and the ratio m:n = 2:3

x = (mx2 + nx1)/(m+n)
x = [2*4 + 3*(-1)]/(2+3)
x = 5/5 =1

y = (my2 + ny1)/(m+n)
y = [2*(-1) + 3*4)/(2+3)
y =10/5 =2

hence the coordinates of the point p(x,y) = P(1,2)

next is to find the gradient of the line AB
m1 = (y2-y1)/(x2-x1)
= (-1-4)/[4-(-1)]
= -5/5 = -1
let the gradient of the line L perpendicular to AB be m2

for any two perpendicular lines, m1*m2 = -1 or m2= -1/m1
m2= -1/-1 = 1

the equation of a straight line given gradient m2 and p(1,2) can also be written as;
yo-y = m2(xo-x)
hence, for line L
yo-2 = 1(xo-1)
yo =xo-1+2
yo = xo+1
OR simply y=x+1
thanks

Richiez:

Question 1
Now we have to rewrite the straight line eqn
x + y = 7
x + y - 7 = 0 ................(1)
For any straight line Ax+By+c=0 and a point (m,n) on same plane, the perpendicular distance between the point and the plane is given by;
d = (lAm+Bn+Cl)/√(A^2 + B^2)
obviously, A=1 , B=1 , C+ -7 , m=4 , n=5
therefore;
d = (l4+5-7l)/√(1^2 +1^2)

d = l2l/√(2) = 1.414units(approx)

Question 2
here, the main task is to find the coordinates of the point on line AB, where the line L cuts it in the ratio 2:3

Let the coordinate of the point be P(x,y)
For the line AB;
A(x1,y1) = A(-1,4)
B(x2,y2) = B(4,-1)

and the ratio m:n = 2:3

x = (mx2 + nx1)/(m+n)
x = [2*4 + 3*(-1)]/(2+3)
x = 5/5 =1

y = (my2 + ny1)/(m+n)
y = [2*(-1) + 3*4)/(2+3)
y =10/5 =2

hence the coordinates of the point p(x,y) = P(1,2)

next is to find the gradient of the line AB
m1 = (y2-y1)/(x2-x1)
= (-1-4)/[4-(-1)]
= -5/5 = -1
let the gradient of the line L perpendicular to AB be m2

for any two perpendicular lines, m1*m2 = -1 or m2= -1/m1
m2= -1/-1 = 1

the equation of a straight line given gradient m2 and p(1,2) can also be written as;
yo-y = m2(xo-x)
hence, for line L
yo-2 = 1(xo-1)
yo =xo-1+2
yo = xo+1
OR simply y=x+1
thanks

Fetus: ....kindly help with dis...simplify tan²(1/4¥ - 1/2€)....where ¥ and € are pie and teetha respectively....tanks

TOSIN ACCA: I have tried solving it yet no solution.I stopped at this point: a-ac+or -sqrt(4a^2c^2-8a^2c+8rc). I guess there is no solution to it. Even the past questions booklet said so thanks for your help. I'll be back with a handful of questions tomorrow.God loves you.

johnpaul1101: plz help me with this: in a two-digit number, the sum of the digits is 8, the difference between this number and the number with the digits reversed is 54. What is this number?

ositadima1:

1) in a two-digit number... "x" and "y"
2) the sum of the digits is 8... x+y=8
3) the difference between this number and the number with the digits reversed is 54... [10y+x]-[10x+y]=54
tnx osita, this was where i got confused
Now,
10y+x-10x-y=54
9y-9x=54
y-x=6

x+y=8
-x+y=6
add these two,
2y=14
y=7
x=1

What is this number?... 71

Freiburger: Prove that Tanx = Sinx/Cosx.

johnpaul1101: a woman's age and her son's add to 45years. Five years ago, the woman was 6times as old as her son, how old was the woman when the son was born?
After solving it, i got 45years which is wrong, plz help

Freiburger:
Answer should be 25

Lets assume the the woman is presently 35 and her son is 10 that adds up to be 45 as stated in the question.
5 years ago the woman was 30, since she is now 35 and her son was 5, since he is now 10. so the woman was 6 times older
so, 5years before the son was born the woman should be 25

@ johnpau1101 how you are O.k with that explanation.

johnpaul1101: this is another similar question:bisi's and fibie's ages add up to 29, 7years ago, bisi was twice as old as fibie, find their present ages.

biolabee: b + f = 29 ---------(1) 7 years go bisi was b - 7 years old fibie was f - 7 However she was twice as old as fibie thus b - 7 = 2(f - 7) expand b - 7 = 2f - 14 b = 2f - 7 ---------(2) substitute in equation 1 (2f-7) + f = 29 3f - 7 = 29 3f = 36 f =12 bisi is thus 29 - 12 = 17 years

Richiez:

3√(10+ 6√3)
the above can be re-written as;
(10+ 6√3)^1/3 = 10^1/3[1+ 0.6√3]^1/3

for fractional index, the binomial series is given by;
(1+x)^1/n = 1 + x/n + 1/n(1/n -1)x^2/2! + 1/n(1/n -1)(1/n -2)x^3/3!+..........
Note that more accuracy is attained when more terms in the series are used
in our case, 1/n=1/3 and x=0.6√3

2.154[ 1 + (1/3)(0.6√3) + 1/3(1/3 -1)(0.6√3)^2/2! + 1/3(1/3 -1)(1/3 -2)(0.6√3)^3/3!]

2.154[ 1 + 0.346 - 0.12 + 0.069 ]

= 2.789 (approx)
remember due to approximations, and the fact that we used only the first three terms, this result will only be approximate to the calculator value...but this is the best method to use in the absence of a calculator
thanks

Ghettoguru: Help with these ones guys.

1. Integrate :

(a) ∫(x³ + 3x² + 5x +1)e^x dx
(b) ∫ln(x² + 2x) dx

2. Find the antiderivatives of the following:

(a) x sec²(x²)tan²(x²)dx
(b) x³4dx/(4 + x²)

Johnpaul88: @Johnpaul1101,
I'm happy that my name sake is here. Congrats for coming here and improving yourself.