1. ABCD of a swimming pool, 12 m long. AB is the horizontal top edge. AD is the depth of the shallow end and BC the depth of the deep end.
i) Calculate the angle that DC makes with the horizontal
ii) Calculate the length of the sloping edge, DC
iii) If the pool is 5m wide, calculate the total surface area.
2.One face of the roof of a house in the shape of a parallelogram ABCD. The angle ABF= 40 degrees and the length AB = 8m. AI represents the rafter placed perpendicular to BC such that FC = 5m.
Calculate i) the length BI
ii) the length AI
iii) the area of ABCD
3. Differentiate the following i) x + 1/x
ii) 2cos x – sin x + 3x^7+ x^4
iii) √(x^3) +11x + 7
b) Integrate 3x^5 – 2sin x dx
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ques 3
ReplyDeletey = x + 1/x
x + x^-1
dy/dx = 1 + -x^-2
ques 3 ii
ReplyDeletey = 2cos x - sin x + 3x^7 + x^4
dy/dx = 2sin x + cos x + 21x^6 + 4x^3
ques 3 iii
ReplyDeletey = (x^3)^0.5 + 11x + 7
x^1.5 + 11x + 7
dy/dx = 1.5x^0.5 + 11
ques 3 b
ReplyDelete3x^5 - 2sin x dx
integral
(3x^6)/6 + 2cos x + C
(x^6)/2 + 2cos x + C
This comment has been removed by the author.
ReplyDelete(3)ii
ReplyDeletey = 2cos x - sin x + 3x^7 + x^4
dy/dx = 2sin x + cos x + 21x^6 + 4x^3
This comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteI think it's impossible to calculate an exact value to answers what's asked for in question 1.i) and ii). Not enough information is given in the problem.
ReplyDeleteyes u can answer it there is enough info. plus check yuh grammer.....
ReplyDeleteok ok, how do I derive a value for the angle DC. You don't know crucial lenghts: AD and BC.
ReplyDeleteBtw, check you spelling
try drawin d diagram and see if it is a little clearer
ReplyDeletesorry fuh d spellin
I've done the drawing, but the only answer I can realistically derive is the surface area for part iii)
ReplyDeletenum(3)(II)
ReplyDelete2cosx-sinx+3x^7+x^4
-2sinx-cosx+21x^6+4x^3
the diff of cosx=-sinx
the diff of sinx= cosx
i agree with Ms_Jones...it is impossible to find an exact value to answer part (i) and (ii)to this question...unless...lol you use variables to represent the missing values.
ReplyDeletesay for example since you don't have a value for BC you let it be represented by x...the value of AD could be found by extending the lenght of the line to equal the lenght of the line BC...when this is done the extenion could by refered to as y...this would now give rise to x-y being the expression for AD...now given all these values the problem can be solved
ReplyDeleteto calculate the angle that DC makes with the horizontal:
ReplyDeletetan (tata) = opposite/adjacent
tan (tata) = y/12
12tan(tata)= y
(tata) = (tan inverse)(y/12)
to calculate the lenght of the sloping edge DC:
ReplyDeletewhen tata = (tan inverse)(y/12)
cos[(tan inverse)(y/12)]= adj/hyp
cos[(tan inverse)(y/12)]= 12/hyp
hyp*cos[(tan inverse)(y/12)]=12
hyp(DC) = 12/cos[(tan inverse)(y/12)]
even though it is ovious one can use variables to greate an express to calculate the answer, i beleive that info. is missin cus the question clearly stated to calculate.
ReplyDelete3. i) x + 1/x
ReplyDeletedy/dx = 1 -1x^-2
3. ii) 2cos x – sin x + 3x^7+ x^4
ReplyDeletedy/dx = 2(-sin x) - cos x + 21x^6 + 4x^3
3. iii) √(x^3) +11x + 7
ReplyDeletey= x^3/2 + 11x + 7
dy/dx = 3/2x^(1/2) + 11
Integral of 3x^5 – 2sin x dx
ReplyDelete>> (2x^6)/6 + 2cos...
im not too sure what the integral of dx is. can someone explain?
for ques 3(¡¡)
ReplyDeletethe diff of cosx=-sinx
the diff of sinx= cosx
2cosx-sinx+3x^7+x^4
-2sinx-cosx+21x^6+4x^3
for ques 3(¡¡)
ReplyDelete3x^5 - 2sin x dx
the integral
(3x^6)/6 + 2cos x + C
(x^6)/2 + 2cos x + C
3)i)
ReplyDeletey = x + 1/x
x + x^-1
dy/dx = 1 + -x^-2
3) ii)
ReplyDelete2cos x – sin x + 3x^7+ x^4
dy/dx = 2(-sin x) - cos x + 21x^6 + 4x^3
3) iii)
ReplyDeletey = (x^3)^0.5 + 11x + 7
x^1.5 + 11x + 7
dy/dx = 1.5x^0.5 + 11
ques 3 iii
ReplyDeletey = (x^3)^0.5 + 11x + 7
x^1.5 + 11x + 7
dy/dx = 1.5x^0.5 + 11
ques 3 ii
ReplyDeletey = 2cos x - sin x + 3x^7 + x^4
dy/dx = 2sin x + cos x + 21x^6 + 4x^3
ques 3 b
ReplyDelete3x^5 - 2sin x dx
integral
(3x^6)/6 + 2cos x + C
(x^6)/2 + 2cos x + C
from question 3b
ReplyDelete3x^5 - 2sin x dx
integral
(3x^6)/6 + 2cos x + C
(x^6)/2 + 2cos x + C
3)(i) x + 1/x
ReplyDeleted/dx= x^1-1 + x^-1-1
d/dx= 1 + x^-2
d/dx= 1 + 1/x^2
3)(ii) 2cosx - sinx + 3x^7 + x^4
ReplyDeleted/dx= -2sinx - cosx + ((3 x 7)x^ 7-1)+ (4 x 1)x^4-1)
d/dx=-2sinx - cosx + 21x^6+ 4x^3
3)(iii) √(x^3)+ 11x +7
ReplyDelete=((x^3)^1/2)+ 11x^1/2 + 7^1/2
=(x^3*1/2)+ 11x^1/2+ 2.65
d/dx= (1.5x^1.5-1)+ (11*1/2x^1/2-1)
d/dx= (1.5x^1/2) + (5.5x^-1/2)
(3) part 1
ReplyDeletey = x + 1/x
x + x^-1
dy/dx = 1 + -x^-2
(3) part 2
ReplyDeletey = 2cos x - sin x + 3x^7 + x^4
dy/dx = 2sin x + cos x + 21x^6 + 4x^3
3) (b) 3x^5 - 2sin x dx
ReplyDelete∫ 3x^5 - 2sinx dx= (3x^5+1)/6 +cosx
∫ (3x^6)/6 + cosx + c
∫ (x^6)/2 + cosx + c
(3) part
ReplyDeletey = (x^3)^0.5 + 11x + 7
x^1.5 + 11x + 7
dy/dx = 1.5x^0.5 + 11
(3) part b
ReplyDelete3x^5 - 2sin x dx
integral
(3x^6)/6 + 2cos x + C
(x^6)/2 + 2cos x + C
also 1/2x^6+2cos x +c
3) y=x+1/x
ReplyDeletey=x+x^-1
dy/dy= 1+x^-2
3ii) y= 2cosx-sinx+3x^7+x^4
ReplyDeletedy/dx= -2sinx+cosx+21x^6+4x^3
3)(i) x + 1/x
ReplyDeleted/dx= x^1-1 + x^-1-1
d/dx= 1 + x^-2
d/dx= 1 + 1/x^2
(ii) 2cosx - sinx + 3x^7 + x^4
d/dx= -2sinx - cosx + ((3 x 7)x^ 7-1)+ (4 x 1)x^4-1)
d/dx=-2sinx - cosx + 21x^6+ 4x^3
(iii) square root(x^3)+ 11x +7
=((x^3)^1/2)+ 11x^1/2 + 7^1/2
=(x^3*1/2)+ 11x^1/2+ 2.65
d/dx= (1.5x^1.5-1)+ (11*1/2x^1/2-1)
d/dx= (1.5x^1/2) + (5.5x^-1/2)
question 3:
ReplyDeletey = x + 1/x
x + x^-1
dy/dx = 1 + -x^-2
dy/dx cos x=sinx
ReplyDeletedy/dx sinx=cosx
y = 2cos x - sin x + 3x^7 + x^4
dy/dx = 2sin x + cos x + 21x^6 + 4x^3
3 (iii)
ReplyDeletey = (x^3)^0.5 + 11x + 7
x^1.5 + 11x + 7
dy/dx = 1.5x^0.5 + 11
3x^5 - 2sin x dx
ReplyDeleteintegrate:
(3x^6)/6 + 2cos x + C
(x^6)/2 + 2cos x + C
3(i)
ReplyDeletex+1/x
x+x^-1
dy/dx=1-x^-2
dy/dx=1-1/x^2
3(ii)
ReplyDelete2cosx-sinx+3x^7+x^4
dy/dx=-2sinx-cosx+21x^6+4x^3
3(iii)
ReplyDelete(√x^3)+11x+7
(x^3)^1/2+11x+7
(x^3/2)+11x+7
dy/dx= 3/2 x^1/2 +11
3(b)
ReplyDelete∫3x^5-2sinx dx
3x^6/6 + 2cosx
x^6/2 + 2cosx
y something really wrong with question one
ReplyDeletequestion one
ReplyDeletethe area of the surface of the pool is
12*5
60m^2
3 i)
ReplyDeletey = x + 1/x
y= x + 1x^-1
dy/dx = 1 - x^-2
dread i dnt understand question #2!!!!!! where did F come from...or how does it fit in to the question....someone please help...:(
ReplyDeleteas for #3
ReplyDeletei)y = x + 1/x
y = x + 1x^-2
dy/dx= 1 - x^-2
=
ReplyDeletequestion 3
y = x + 1/x
x + x^-1
dy/dx = 1 + -x^-2
question 3 ii
ReplyDelete2cosx-sinx+3x^7+x^4
dy/dx=-2sinx-cosx+21x^6+4x^3
question 3 (iii) √(x^3)+ 11x +7
ReplyDelete=((x^3)^1/2)+ 11x^1/2 + 7^1/2
=(x^3*1/2)+ 11x^1/2+ 2.65
d/dx= (1.5x^1.5-1)+ (11*1/2x^1/2-1)
d/dx= (1.5x^1/2) + (5.5x^-1/2)
ques 3
ReplyDeletey = x + 1/x
x + x^-1
dy/dx = 1 + -x^-2
ques 3 ii
ReplyDeletey = 2cos x - sin x + 3x^7 + x^4
dy/dx = 2sin x + cos x + 21x^6 + 4x^3
ques 3 iii
ReplyDeletey = (x^3)^0.5 + 11x + 7
x^1.5 + 11x + 7
dy/dx = 1.5x^0.5 + 11
ques 3 b
ReplyDelete3x^5 - 2sin x dx
integral
(3x^6)/6 + 2cos x + C
(x^6)/2 + 2cos x + C
3)(i) x + 1/x
ReplyDeleted/dx= x^1-1 + x^-1-1
d/dx= 1 + x^-2
d/dx= 1 + 1/x^2
(ii) 2cosx - sinx + 3x^7 + x^4
d/dx= -2sinx - cosx + ((3 x 7)x^ 7-1)+ (4 x 1)x^4-1)
d/dx=-2sinx - cosx + 21x^6+ 4x^3
(iii) square root(x^3)+ 11x +7
=((x^3)^1/2)+ 11x^1/2 + 7^1/2
=(x^3*1/2)+ 11x^1/2+ 2.65
d/dx= (1.5x^1.5-1)+ (11*1/2x^1/2-1)
d/dx= (1.5x^1/2) + (5.5x^-1/2)
3)(i) x + 1/x
ReplyDeleted/dx= x^1-1 + x^-1-1
d/dx= 1 + x^-2
d/dx= 1 + 1/x^2
(ii) 2cosx - sinx + 3x^7 + x^4
d/dx= -2sinx - cosx + ((3 x 7)x^ 7-1)+ (4 x 1)x^4-1)
d/dx=-2sinx - cosx + 21x^6+ 4x^3
(iii) square root(x^3)+ 11x +7
=((x^3)^1/2)+ 11x^1/2 + 7^1/2
=(x^3*1/2)+ 11x^1/2+ 2.65
d/dx= (1.5x^1.5-1)+ (11*1/2x^1/2-1)
d/dx= (1.5x^1/2) + (5.5x^-1/2)
for ques 3(III)
ReplyDeletey = (x^3)^0.5 + 11x + 7
x^1.5 + 11x + 7
dy/dx = 1.5x^0.5 + 11