10. If Integrate (4x + k ) wrt x with 2 and 1 as bounds = 1 find k
11. Find the area of the region enclosed by y = 4 – x ^2 and the x –axis
from x = -1 to x = 1
12. The amount of liquid V cm ^3 in a leaking tank at time t secs. is given
by V = ( 20 – t ) ^3 for 0 ≤ t ≤ 20
Find the rate at which water leaves the tank when t = 5 secs.
13. Given 3y –x + 6 = 0
(1) Make y the subject of the formula 3x - y + 6 - 0
State the gradient, and the y – intercept
14. Using any suitable method , solve the quadratic equation
2x^2 – 4x + 1 = 0
( give answers to 3 s.f )
c) The floor of a room is in the shape of rectangle . The floor is ‘ C ‘ metres long. The width is 2 metres less tha its length .
(1) State in terms of C
(a) the width of the floor
(b) the area of the floor
(ii ) If the area of the floor is 15 m ^2 write an equation in C to show the information.
(iii) Use the equation to determine the width of the floor
15 Solve for p and r given
3p + 2r = 7
p2 – 2r = 11
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10. the integration of 4x+k is 4x2/2+kx=1
ReplyDeletebetween the boundaries of 2 and 1
[2(2)2+k(2)]-{2(1)2+k(1)]=1
[2(4)+k(2)]-[2+k]=1
[8+2k]-[2+k]=1
8-2+2k+k=1
6+3k=1
3k=1-6
3k=5
k=5/3
since the region is between -1 and 1/
ReplyDeleteso that there is no complications i would seperate it into two sections.
therefor the regions would be between -1 and 0
and 0 and 1.
the integratio of 4-x^2 is
4x-x^3/3
so between -1 and o the area is
[4(0)-(0)^3/3]-[4(-1)-(-1)^3/3)
=11/3 units^2
for the second section
0 and 1
[4(0)-(0)^3/3]-[4(1)-(1)^3/3]
=11/3units^2
so total area=11/3+11/3
=22/3 units^2
12. integrate v to get rate and to fine the rate at 5 sec sub t=s in the integration.
ReplyDelete(20-t)^3
integration=
(20-t)^4/80
therefor at t=5
=
(20-5)^4/80
(15)^4/80
50624/80
=632.81
13.
ReplyDelete3y-x+6=0
3y=x-6
y=1/3x-6/3
y=1/3x-2
the grad=1/3
the y-intercept is at x=0
y=1/3(0)-2
y=2
the y-intercept is 2
14. the quadratic could not of been factorized . si i used the quadratic equation
ReplyDelete-b+/-(sqrt(-b2-4ac)/2a
where b=-4
a=2
c=1
therefore
=-(-4)+/-(sqrt((-4)2-4(2)(1))/2(2)
=4+/-(sqrt(16-8)/4
=4+/-(2.828)/4
therefore
=4+(2.828)/4 and =4-(2.828)/4
=1.707 =0.293
therefore x=1.707 and=0.293
length = c
ReplyDeletewidth=c-2
area= length * width
area=(c)(c-2)
area= c^2-2c
since the area =15
therefore
c^2-2c=15
c^2-2c-15=0
factorizing
(c+3)(c-5)
therefore
c= -3 or 5
when c=-3 the area is not 15 therefore c is not -3
c=5 units
length =5units
width=5-2=3 units
3p+2r=7....(1)
ReplyDelete2p-2r=11...(2)
from 1
2r=7-3p....(3)
sub 3 into 2
2p-(7-3p)=11
2p-7+3p=11
2p+3p=11+7
5p=18
p=18/5
p=3.6
sub p into 1
3(3.6)+2r=7
10.8 +2r=7
2r=7-10.8
2r=-3.8
r=-3.8/2
r--1.9
therefore r=-1.9 and p=3.6
question 15.
ReplyDelete3p + 2r = 7
p2 – 2r = 11
solving simultaneously
adding them together
3p+2p+ 2r+-2r=7+11
5p+0=18
5p=18
p=18/5=3.6
subbing p=3.6
into
3p + 2r = 7
3(3.6)+2r=7
10.8+2r=7
2r=7-10.8
2r=-3.8
r=-3.8/2
r=-1.9
This comment has been removed by the author.
ReplyDeleteQuestion 14
ReplyDeletea) The width is
w=(c-2) where c is the length,and w is 2 times smaller than l
b) Area=l*w
=c*(c-2)
=c^2-2c
14
ReplyDeleteii) Area is 15m^2
Area=15m^2
then 15= C^2-2c (this is known after concluded
therefore 15+2c-c^2=0
or c^2-2c-15=0
factorizing this we get
(c-5)(c+3)=0
c=5 & c=-3
w=(c-2)
substituting for c
W=(5-2)
=3
or
w=(-3-2)
=-5
15)
ReplyDelete3p+2r=7
p2-2r=11
*1st eq.by 2
6p+4r=14
*2nd eq.by 3
6p-6r=33
Note- this is an elimination process (observe below)
(-) equation 1 from equation 2
we get
10r=47
r=47/10
r=4.7
Finding for p now
Note You can Choose any equation u started with
3p+2r=7
3p+2(4.7)=7
P=7-9.4/3
p=-0.8
green latern are u sure number 10 correct?
ReplyDelete15). 3p+2r= 7.............eq (1)
ReplyDelete2p-2r= 11............eq (2)
add eq (1) and eq (2).
5p= 18
p= 18/8
p= 3.6
subst in eq (1)
when p= 3.6
3(3.6)+2r= 7
10.8+ 2r= 7
2r= 7-10.8
2r= -3.8
r= -3.8/2
r= -1.9
therefore p= 3.6 and r= -1.9
13).
ReplyDelete3y-x-6= 0
3y= x-6
y= x/3-6/3
y= 1/3x-2
therefore gradient= 1/3 and y-intercept= -2
14)
ReplyDelete2x^2-4x+1= 0
ax^2+bx+c= 0
therefore a=2, b= -4, c= 1
x= -b+-sqrt b^2-4ac/2a
x= -(-4)+-sqrt (-4)^2-4(2)(1)/2(2)
x= 4+-2.828/4
x= 4+2.828/4 OR x= 4-2.828/4
x= 1.707 OR x= 0.293
lenght= c
ReplyDeletewidth= c-2
area= c(c-2)
when area is 15 m^2
15= c(c-2)
15= c^2-2c
c^2-2c-15= 0
c^2+3c-5c-15= 0
c(c+3)-5(c+3)= 0
(c+3)(c-3)= 0
therefore c= -3 or c= 5
when c=-3 the area is not 15m^2, so when c= 5 the area is 15m^2.
so therefore lenght= 5 m
width= 5-2
= 3 m
14) 2x^2 - 4x +1=0
ReplyDeletex= (-b +/- √b^2 - 4(a)(c))/ 2a
x= ( 4 +/- √ 4^2 - 4(2)(1))/2(2)
x= ( 4 +/- √ 16 -8)/4
x= ( 4 +/- √8)/4
x= (4 + √8)/4
x=1.707
x= (4 - √8)/4
x= 0.293
so the resultant values are x= 1.707 and x= 0.293
This comment has been removed by the author.
ReplyDelete14(1)(b) Area = Width * Length
ReplyDeletesub width = 2 - length
Area = (2 – length) * length
sub length= c
Area =(2 – c)* c
Area= 2c - c^2
14)(a)width= 2 - length
ReplyDeletetherefore if length is c
the width in terms of c is:
width= 2 - c
10)
ReplyDeletethe integral of 4x+k is 4x2/2+kx=1
between the limits of 2 and 1
[2(2)2+k(2)]-{2(1)2+k(1)]=1
[2(4)+k(2)]-[2+k]=1
[8+2k]-[2+k]=1
8-2+2k+k=1
6+3k=1
3k=1-6
3k=5
k=5/3
not sure if it is correct though
ReplyDelete13)
ReplyDelete3y-x-6= 0
3y= x-6
y= x/3-6/3
y= 1/3x-2
therefore gradient= 1/3 and y-intercept= -2
c)
ReplyDelete(1)(a) W = C-2 m
(b) A = C^2 -2C
(ii) C^2 - 2C = 15m^2
ReplyDelete(iii)
ReplyDeleteA = C^2 - 2C = 15 m^2
C^2 - 2C = 15
making this into a quadratic;
C^2 - 2C -15 = 0
C^2 -5C + 3C -15 = 0
C(C-5) +3(C-5)
(C-5) (C+3)
C = 5 [since a length cannot be -ve]
C = 5
ReplyDeleteW = C - 2 m
W = (5) - 2 m
W = 3 m
13)
ReplyDelete3x-y+6=0
y=3x+6
which is of the form y=mx+c
where the gradient(m) =3
and the intercept = 6
15)
ReplyDelete3p+2r=7.....(1)
2p-2r=11....(2)
(1)+(2)
5p=18
p=18/5
sub into eqn (1)
3(18/5)+2r=7
54/5+2r=7
54+10r=35
10r=-19
r=-19/10
12)
ReplyDeleteV= (20-t)^2
V= (20-t)(20-t)(20-t)
V= (t^2-40t+400)(20-t)
V= 60t^2-1200t+8000-t^3
dv/dt= 120t-1200-3t^2
when t= 5 secs.
dv/dt= 120(5)-1200-3(5)^2
= 600- 1200- 75
= -675
so therefore the rate at which water leaves the tank is -675.
11).
ReplyDeletey= 4-x^2
the integral of that is 4x-x^3/3
now using the bounds of -1 to 1
[4(1)-(1)^3/3]-[4(-1)-(-1)^3/3]
[4-1/3]-[-4+1/3]
[11/3]-[-11/3]
11/3+11/3
22/3
so therefore the area under the curve 22/3units^2
10).
ReplyDelete4x+k wrt x
the integral of that is 4x^2/2+kx
= 2x^2+kx
now using bounds 2 and 1 and that equal to 1
[2(2)^2+k(2)]-[2(1)^2+k(1)]=1
8+2k-2-k= 1
k+6=1
k= 1-6
k= -5
so therefore k= -5
question 1
ReplyDeletethe integral of 4x+k is 4x2/2+kx=1
between the limits of 2 and 1
[2(2)2+k(2)]-{2(1)2+k(1)]=1
[2(4)+k(2)]-[2+k]=1
[8+2k]-[2+k]=1
8-2+2k+k=1
6+3k=1
3k=1-6
3k=5
k=5/3
question 15).
ReplyDelete3p+2r= 7.............eq (1)
2p-2r= 11............eq (2)
add eq (1) and eq (2).
5p= 18
p= 18/8
p= 3.6
subst in eq (1)
when p= 3.6
3(3.6)+2r= 7
10.8+ 2r= 7
2r= 7-10.8
2r= -3.8
r= -3.8/2
r= -1.9
therefore p= 3.6 and r= -1.9
question 14)
ReplyDelete2x^2-4x+1= 0
ax^2+bx+c= 0
therefore a=2, b= -4, c= 1
using quadratic formula
x= -b+-sqrt b^2-4ac/2a
x= -(-4)+-sqrt (-4)^2-4(2)(1)/2(2)
x= 4+-2.828/4
x= 4+2.828/4 OR x= 4-2.828/4
x= 1.707 OR x= 0.293
V = ( 20 – t ) ^3
ReplyDeleteWhen t=5
V = ( 20 - (5)) ^3
= (15)^3
= 3375
WoW..that looking weird...is that correct???
3x - y + 6 - 0
ReplyDeleteI'll take it that the '-' before the 0 was supposed to be an '='.
3x - y + 6 = 0
making y the subject:
3x + 6 - y = 0
y = 3x + 6
y = 3x + 6 is in the form y = mx + c where 'm' is the gradient and 'c'is the y-intercept. therefore:
ReplyDeletegradient = 3
y-intercept = 6
i'll take it that the 'p2' supposed to be '2p'
ReplyDelete15) Solve for p and r given
3p + 2r = 7...eq(1)
2p – 2r = 11...eq(2)
eliminate r by adding the two equations
5p = 18
p = 3.6
Susbst. p = 3.6 into eq (1)
3(3.6) + 2r = 7
10.8 + 2r = 7
2r = =3.8
r = -1.9
therefore p = 3.6
r = -1.9
number (10)
ReplyDeletethe integration of 4x+k is 4x2/2+kx=1
between the boundaries of 2 and 1
[2(2)2+k(2)]-{2(1)2+k(1)]=1
[2(4)+k(2)]-[2+k]=1
[8+2k]-[2+k]=1
8-2+2k+k=1
6+3k=1
3k=1-6
3k=5
k=5/3
number (11)
ReplyDeletey= 4-x^2
the integral of that is 4x-x^3/3
now using the bounds of -1 to 1
[4(1)-(1)^3/3]-[4(-1)-(-1)^3/3]
[4-1/3]-[-4+1/3]
[11/3]-[-11/3]
11/3+11/3
22/3
so therefore the area under the curve 22/3units^2
number (12)
ReplyDeleteV= (20-t)^2
V= (20-t)(20-t)(20-t)
V= (t^2-40t+400)(20-t)
V= 60t^2-1200t+8000-t^3
dv/dt= 120t-1200-3t^2
when t= 5 secs.
dv/dt= 120(5)-1200-3(5)^2
= 600- 1200- 75
= -675
so therefore the rate at which water leaves the tank is -675.
number (13)
ReplyDelete3x-y+6=0
y=3x+6
which is of the form y=mx+c
where the gradient(m) =3
and the intercept = 6
number (15)
ReplyDelete3p+2r=7.....(1)
2p-2r=11....(2)
(1)+(2)
5p=18
p=18/5
sub into eqn (1)
3(18/5)+2r=7
54/5+2r=7
54+10r=35
10r=-19
r=-19/10
. 3p + 2r = 7
ReplyDeleteP^2 – 2r = 11
Equation (1) add equation (2)
3p + p^2 = 18
P + p^2 = 18/3
P + p^2 = 6
Cannot complete
HELP!!!!!!!
13. Given 3y –x + 6 = 0
ReplyDelete(1) Make y the subject of the formula
3y - x + 6 = 0
3y - x = -6
3y = -6 + x
y = (-6 + x)/3
y = -2 + x/3
(a) the width of the floor in terms of 'c'
ReplyDeletew = c - 2
(b) the area of the floor in terms of 'c'
ReplyDeletearea = L * W
L = C
W = C -2
area = C(C-2)
c) The floor of a room is in the shape of rectangle . The floor is ‘ C ‘ metres long. The width is 2 metres less tha its length .
ReplyDelete(1)
(a) the width of the floor---w = C-2
(b) the area of the floor----A = C(C-2)
A = C^2 - 2C
(ii )area = 15 m ^2
15 m ^2 = C^2 - 2C
(iii) Use the equation to determine the width of the floor
(ii )C^2 - 2C =15 m ^2
w = C-2
C=w+2
sub C=w+2 into the area
w+2
(w+2)^2 - 2( w+2) =15
w^2 +4w+4 -2w -4 = 15
w^2 + 2w -15 = 0
(w+5)(w-3)
width=3
15)
ReplyDelete3p + 2r = 7---------(1)
2p – 2r = 11--------(2)
equation (1)
2r = 7 -3p----(3)
sub (3) into (2)
2p - (7 - 3p)= 11
5p = 18
p=18/5
sub p into eq'n (3)
2r = 7 -3(18/5)
2r = 7 - 10.8
r = -3.8/2
r= 1.9
If Integrate (4x + k ) wrt x with 2 and 1 as bounds = 1 find k
ReplyDeleteintegral of (4x + k)
= [4x^(1+1)]/2 + kx
= 4x^2/2 + kx
= 2x^2 + kx
then find the value of k by substituting the bounds into the equation:
ie. [2(2^2) + k(2)]-[2(1^2) + k(1)] = 1
[8 + 2k]- [2 + k] = 1
6 + 3k = 1
3k = 1 - 6
3k = -5
k = -5/3
11. Find the area of the region enclosed by y = 4 – x ^2 and the x –axis
ReplyDeletefrom x = -1 to x = 1
integrate the equation y = 4 - x^2
= 4x - x^(2+1)/3 + c
= 4x - x^3/3 + c
then sub. into the following equation to find the area....
area = [4(1) - (1^3)/3]-[4(-1) - (-1^3)/3]
area = 3.67 - (-3.67)
area = 7.34
I have to say out of all the blogs on this particular page, 'St. Anns graduate' made the most sense...I totally agree with his/her style of working out problems...Green latern comments were also very good, he made a few minor mistakes however his approach to the questions is commendable!
ReplyDeleteQuestion 10.
ReplyDelete∫(4x+k) dx
4x^3/3 +kx + c {where c is a constant}
2x^2+ kx + c
For bounds 2 and 1
[2 (2)^2 + k (2)] – [2 (1)^2 + k (1)] =1
[8 + 2k] – [2 +k]= 1
8+ 2k -2 – k = 1
6 + k = 1
k = 1- 6
k = -5
Question 11.
ReplyDeletey= 4-x^2
∫(4-x^2) dx
4x – x^3/3
For Boundaries 1 and -1
[4 (1) – (1)^3/3] – [4 (-1) – (-1)^3/3]
[4- 1/3] – [ -4 + 1/3]
3 2/3 + 3 2/3
7 1/3
Question 12.
ReplyDeletev = (20 - t)^3
v= (400- 40t + t^2) (20-t)
v=8000 – 800t + 20t^2 – 400t + 40t^2 – t^3
v= -t^3 +60t^2 – 1200t + 8000
rate of change is dv/dt
dv/dt = -3t^2 + 120t -1200
at t=5
= -3(5)^2 +120(5) -1200
= -75 + 600 - 1200
= - 675
Question 13.
ReplyDelete3y-x+6-0=0
3y=x-6
y= (x-6)/3
y=1/3x – 2
gradient is dy/dx
dy/dx = 1/3
therefore gradient of line is 1/3
y-intercept occurs when x= 0
y= 1/3(0)- 2
y=2 is where the line cuts the y axis
Question 14.
ReplyDelete2x^2 – 4x +1 =0
Solve equation using Quaratic formula
I have done the working on paper but it seems
Very difficult to type it up
x= 1.71 and x= 0.293 to 3 sig. figs
Question 15.
ReplyDeleteFloor of room is rectangular.
i (a) if Length L = c
Then width W = c-2
(b) Area (A) of rectangle = LxB
A= c(c-2)
A= c^2 – 2c
Question 15. Second part
ReplyDelete(ii) When A= 15m^2, sub A into equation to find c
15= c^2 – 2c
c^2 – 2c – 15= 0
(x -5) (x+3)= 0
X= 5 and x= -3
Since length cannot be negative
Length of floor is 5m
And Width is
W= 5-2
W= 3m
Question 16.
ReplyDelete3p + 2r = 7 ---------(1)
p2 -2r= 11----------(2)
Solving using substsitution method
From equation (1)
r= (7-3p)/2 ---------(3)
Sub (3) into (2)
2p – 2[ (7-3p)/2] = 11
2p – 7 +3p = 11
5p = 11+7
p= 18/5
Sub p= 18/5 into equation (3)
r= (7-3(18/5)/2
r= 7/2 – (54/4 x 1/2)
r = 7/2 – 27/5
r = -19/10
Solved p= 18/5 and r = -19/10