- A rectangle is to have a perimeter 26cm and a length x cm. If the width x equals 3cm, find the length and hence the Area of the rectangle. Find dA/dx and hence find the width of the rectangle giving the maximum area.
- Solve for p and r given 3p + 2r = 7 and p^2 - 2r = 11
- Solve 8x^2 + 3y^2 = 50 and 2x + y = 5
- Write the expression 9x^2 - 9x + 1 in the form a(x + b)^2 + c, where a,b and c are constants. Hence state whether the function y = 9x^2 - 9x + 1 has a maximun or minimum value. State the value of x at which this maximum or minimum value occurs
- EFGH is a parallelogram with EF = 6cm EH = 4.2cm, and angle FEH = 70 degrees. Calculate the length of HF. Calculate the area of the parallelogram EFGH.
- A vertical tover FT has a vertical antenna TW mounted on top of the tower. A point P is on the same horizontal ground as F such that PF = 28 m and the angle of elevation of T and W from P are 40 and 54 resp. Calculate the length of the antenna TW.
- A vertical pole AD and a vertical tower BC stands on horizontal ground XABY. The height of the pole is 2.5 m and the angle of depressionn of B from D is 15 degrees abd the angle of elevation of C from D is 20 degrees. DE is a horizontal line. Calculate AB and the height of the tower BC.
- Find the coordinates of the stationary points on the graph of y = x^3 - 12 x - 12.
- Factorise x^2 + x -12
- Evaluate (5x^2 + 4x + 1) + (-7x + 2)
- Factorise 5x^2 + 13x - 6
- Simplify (8x^4 - 2x^2)/2x^2
- Factorise 3x^2 - 48
- KNM is a right angle triangle with KNL also being a right angle triangle. AN = 6 cm and NM = 15.6 cm and angle KLN is 52 degrees. KLM is a straight line. Calculate the size of angle KMN and the length LM
- F(x) = 3x^2 - 12x + 5. Write in the form a(x + b)^2 + c, where a,b and c are constants. Hence determine the minimum value of f(x) and the coordinates of the minimum point.
- y = 6x^2 + 32/x^3, find dy/dx
- y = x^2 + 54/x, find dy/dx
- y = 1 + 4x^3, find dy/dx
- Find the coordinates of the stationary points on the graph of y = x^2 + x^3.
- The curve y = 27 - x^2 has the points P and S on the curve. The point R and Q lie on the x-axis and PQRS is a rectangle. The length of OQ is t units. Find the length of PQ in terms of t and show the area of PQRS is A = 54t - 2t^3. Find the value of t for which A has a stationary value. What is the stationary value of A and determine its nature (max or min)
Thursday, December 3, 2009
Questions
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QUESTION 1
ReplyDeletewell if the perimeter = 26
2L + 2W = 26
BUT W=3CM
2L + 2(3) = 26
2L=26-6
2L=20
L=10cm
QUESTION 1 CONTD
ReplyDeleteAREA = L x W = 10 x 3=30cm^2
5) e^2=h^2+f^2-2hfcosx
ReplyDeletee^2=(6)^2+(4.2)^2-2(6)(4.2)cos70
e^2=36+17.64-50.4(0.34)
e^2=53.64-17.136
e=squar root of 36.504
e=6.04
i.e the length of HF = 6.04
The area of a parallelogram = base * hight
ReplyDeletethere is no perpendicular hight so area can not be calculated
6)Triangle 1 (PWF)
ReplyDeletetan tata=opp/adj
i.e opp=adj tan tata
opp=28 tan54
opp=28(1.38)
opp=38.64
i.e FW=38.64
6)Triangle 2 (PTF)
ReplyDeletetan tata=opp/adj
i.e opp=adj tan tata
opp=28 tan40
opp=28(0.84)
opp=23.52
i.e FT=23.52
6)i.e TW=FW-FT
ReplyDeleteTW=38.64-23.52
TW=15.12
7)Triangle (ABD)
ReplyDeletetan tata=opp/adj
i.e opp=adj tan tata
opp=2.5 tan75
opp=2.5(3.73)
opp=8.4
i.e AB=8.4
This comment has been removed by the author.
ReplyDeletenote
ReplyDeleteAB=8.4
i.e DE=8.4
note
ReplyDeleteAD=2.5
i.e BE=2.5
7)Triangle (CDE)
ReplyDeletetan tata=opp/adj
i.e opp=adj tan tata
opp=8.4 tan20
opp=8.4(0.36)
opp=3.0
i.e EC=3.0
i.e BC=BE+EC
ReplyDeleteBC=2.5+3.0
BC=5.5
y=x^3-12x-12
ReplyDeleteusing loop
dy/dx=3x^2-12
for stationary point dy/dx=o
i.e 3x^2-12=0
3x^2=12
x^2=12/3
x^2=4
i.e x=squar root of 4
x=+or-2
d^2y/dx^2=3x^2-12
ReplyDeleteusing loop
d^2y/dx^2=6x
d^2y/dx^2=6/x
sub when x=2
d^2y/dx^2=6/2
=3
sub when x=-2
d^2y/dx^2=6/-2
=-3
(9) x^2 + x -12
ReplyDeletex^2 + 4x -3x -12
x(x + 4)-3(x +4)
(x+ 4)(x-3)
(8) (5x^2 + 4x + 1) + (-7x + 2)
ReplyDelete5x^2 + 4x + 1 -7x + 2
5x^2 - 3x + 3
Number 1
ReplyDeletegiven that width=3 and perimeter=26, therefore:
P=2L+2W
26=2L+2(3)
26=2L+6
26-6=2L
2L=20
L=20/2
L=10
Number 1 (ii)
ReplyDeletethe area:
A=L*W
A=10*3
A=30cm^2
for part one of number one i forgot the units which is "cm". i don't want to delete and then type over. sorry.
ReplyDeleteQuestion 2
ReplyDeleteif i'm wrong please correct this question
solve for p
3p+2r=7
3p=7-2r
3p/3=7-2r/3
p=7-2r/3
solve for r
3p+2r=7
-2r=3p-7
-2r/-2=3p-7/-2
solve for p
p^2-2r=11
p^2=11+2r
p=11+2r^-2
solve for r
p^2-2r=11
2r=p^2-11
2r/2=p^2-11/2
r=p^2-11/2
For Ouestion no.5, how to find the area of a parellelogram and the diagonal side of HF?
ReplyDeleteQuestion 9
ReplyDeletex^2+x-12
x^2-3x+4x-12
(x^2-3x)(+4x-12)
x(x-3)+4(x-3)
(x-3)(x+4)
Question 11
ReplyDelete5x^2+13x-6
5x^2+15x-2x-6
(5x^2+15x)(-2x-6)
5x(x+3)-2(x+3)
(x+3)(5x-2)
Perimeter=2L + 2b
ReplyDelete=2L + 2(x)
x=3 -->
26=2l + 2(3)
2L=26 - 6
L=20/2
L=10 cm
In finding L.....
ReplyDeleteL=10cm
b= 3cm
Area = L*b
= 10*3
= 30 cm^2
perimeter=26cm
ReplyDeletel=xcm
b=3cm
perimeter of rectangle=2L+2b
26=2L+2(3)
26-6=2L
20=2L
hence L=10cm
area of perimeter=L*b
ReplyDelete3*10=30cm^2
e^2=h^2+f^2-2hfcosx
ReplyDeletee^2=(6)^2+(4.2)^2-2(6)(4.2)cos70
e^2=36+17.64-50.4(0.34)
e^2=53.64-17.136
e=squar root of 36.504
e=6.04
length of HF = 6.04
9)x^2+x-12
ReplyDeletex^2-3x+4x-12
(x^2-3x)(+4x-12)
x(x-3)+4(x-3)
(x-3)(x+4)
7)
ReplyDeleteTriangle (CDE)
tan tata=opp/adj
opp=adj tan tata
opp=8.4 tan20
opp=8.4(0.36)
opp=3.0
hence EC=3.0
6
ReplyDeleteTriangle 1 (PWF)
tan tata=opp/adj
hence opp=adj tan tata
opp=28 tan54
opp=28(1.38)
opp=38.64
i.e FW=38.64
This comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
ReplyDelete2L + 2W = 26
ReplyDeleteW=3CM
2L + 2(3) = 26
2L=26-6
2L=20
L=10cm
area = L x W = 10 x 3=30cm^2
e^2=h^2+f^2-2hfcosx
ReplyDeletee^2=(6)^2+(4.2)^2-2(6)(4.2)cos70
e^2=36+17.64-50.4(0.34)
e^2=53.64-17.136
e=square root of 36.504
e=6.04
length of HF = 6.04
7)Triangle (ABD)
ReplyDeletetan tayta=opp/adj
i.e opp=adj tan tayta
opp=2.5 tan75
opp=2.5(3.73)
opp=8.4
i.e AB=8.4
5x^2+13x-6
ReplyDelete5x^2+15x-2x-6
(5x^2+15x)(-2x-6)
5x(x+3)-2(x+3)
(x+3)(5x-2)
^^question #11
need help with ques 15!!!!!
ReplyDelete1. 2L + 2W = 26
ReplyDelete2L + 2(3) = 26
2L = 26 -6
2L = 20
therefore, L = 20/2
L =10 cm
area = L * W
ReplyDeletearea = 20* 3
area = 30cm^2
P = 26 cm
ReplyDeleteL = x cm
W = 3 cm
P = 2L + 2B or (L + B)*2
26 = 2x + 2*(3)
26 = 2x + 6
26 - 6 = 2x
2x = 20
x = 10cm
therefore since L = x
L = 10cm
1 cont...
ReplyDeleteA = L * B
A = 10 * 3
A = 30cm^2
A = L * B
A = x * 3Aa = 3x
dA/dx = 3
2. 3p + 2r = 7_______1
ReplyDeletep^2 - 2r = 11_____2
eq1+eq2: 3p + p^2 + 2r + - 2r = 18
3p + p^2 = 18
p^2 + 3p - 18 = 0
p^2 + 6p - 3p - 18 = 0
p(p + 6) - 3(p + 6) = 0
(p + 6)(p - 3) = 0
p + 6 = 0 p - 3 = 0
p = -6 p = 3
subst values for p into equation 1:
3p + 2r = 7
3(-6) + 2r = 7
-18 + 2r = 7
2r = 25
r = 12.5
3p + 2r = 7
3(3) + 2r = 7
9 + 2r = 7
2r = -2
r = -1
so corresponding values are:
(p= -6,r=12.5) (p=3,r= -1)
3. 8x^2 + 3y^2 = 50___1
ReplyDelete2x + y = 5____________2
2: 2x + y = 5
y = 5 - 2x____________3
subst eq3 into eq1 : 8x^2 + 3(5 - 2x)^2 = 50
8x^2 + 3(25 - 20x + 4x^2) = 50
8x^2 + 75 - 60x + 12x^2 = 50
20x^2 - 60x + 75 - 50 = 0
20x^2 - 60x + 25 = 0
/5: 4x^2 - 12x + 5 = 0
4x^2 - 2x - 10x + 5 = 0
2x(2x - 1) - 5 (2x - 1) = 0
(2x - 1) (2x - 5) = 0
2x - 1 = 0 2x - 5 = 0
2x = 1 2x = 5
x = 1/2 x = 5/2
subst x values into eqn 2
2x + y = 5
2(1/2) + y = 5
1 + y = 5
y = 4
2x + y = 5
2(5/2) + y = 5
5 + y = 5
y = 0
(x = 1/2,y = 4) (x = 5/2,y = 0)
4. 9x^2 - 9x + 1 = 0
ReplyDelete9x^2 - 9x = -1
/9: x^2 - x = -1/9
add half square of the term in x, i.e. half the square coefficient of x:
x^2 - x + (-1/2)^2 = -1/9 + (-1/2)^2
(x - 1/2)^2 = 5/36
(x - 1/2)^2 - 5/36 = 0
since we previously divided by 9 we must now multiply by 9 to bring back the equation to the original form
x9: 9 [(x - 1/2)^2 - 5/36 = 0]
9(x - 1/2)^2 - 5/4 = 0
the equation is now in the form of a(x + b)^2 + c
since c = -5/4, it is a min value
9(x - 1/2)^2 = 5/4
(x - 1/2)^2 = 5/36
(x - 1/2) = +√5/36 or -√5/36
x - 1/2 = √5/36 or x - 1/2 = -√5/36
x = √5/36 + 1/2 or x = -√5/36 + 1/2
It's best to leave the x-values in this form since they are more accurate than decimals.
if the perimeter = 26
ReplyDelete2L + 2W = 26
BUT W=3CM
2L + 2(3) = 26
2L=26-6
2L=20
L=10cm
area therefore = L x W = 10 x 3
ReplyDelete=30cm^2
11.5x^2+13x-6
ReplyDelete5x^2+15x-2x-6
(5x^2+15x)(-2x-6)
5x(x+3)-2(x+3)
(x+3)(5x-2)
1)
ReplyDeletewidth (W)=3
perimeter (p)=26
P=2L+2W
26=2L+2(3)
26=2L+6
26-6=2L
2L=20
L=20/2
L=10
1)ii)
ReplyDeleteL=10cm
b= 3cm
therefore:
Area = Lxb
= 10x3
= 30 cm^2
2)3p+2r=7
ReplyDeletep^2-2r=11
using elimination method
3p-p^2=-4
-p^2+3p+4=0 *-1
p^2-3p-4=0
p^2-4p+p-4=0
p(p-4)+1(p-4)=0
(p-4)(p+1)=0
p=4 , p=-1
subs.p into 3p+2r=7
p=4
3(4)+2r=7
2r=-5
r=-5/2
p=-1
3(-1)+2r=7
2r=10
r=5
2)
ReplyDelete3p + 2r = 7.........eq1
p^2 - 2r = 11.......eq2
eq1 + eq2
3p + p^2 + 2r + - 2r = 18
3p + p^2 = 18
p^2 + 3p - 18 = 0
p^2 + 6p - 3p - 18 = 0
p(p + 6) - 3(p + 6) = 0
(p + 6)(p - 3) = 0
p + 6 = 0 p - 3 = 0
p = -6 p = 3
substitute p values into eq1:
3p + 2r = 7
3(-6) + 2r = 7
-18 + 2r = 7
2r = 25
r = 12.5
3p + 2r = 7
3(3) + 2r = 7
9 + 2r = 7
2r = -2
r = -1
(p= -6,r=12.5)and when(p=3,r= -1)
3)
ReplyDelete8x^2+3y^2=50
2x+y=5
y=5-2x subs.into 8x^2+3y^2=50
8x^2+3(4x^2-20x+25)=50
20x^2-60x+25=0/5
4x^2-12x+5=0
4x^2-2x-10x+5=0
2x(2x-1)-5(2x-1)=0
(2x-1)(2x-5)=0
x=1/2 , x=5/2
subs. x into 2x+y=5
2(1/2)+y=5
y=4
and
2(5/2)+y=5
y=0
3)
ReplyDelete8x^2 + 3y^2 = 50....eq1
2x + y = 5..........eq2
make y subject of the formula:
2x + y = 5
y = 5 - 2x..........eq3
substitute eq3 in eq1
8x^2 + 3(5 - 2x)^2 = 50
8x^2 + 3(25 - 20x + 4x^2) = 50
8x^2 + 75 - 60x + 12x^2 = 50
20x^2 - 60x + 75 - 50 = 0
20x^2 - 60x + 25 = 0
divide by (5)
4x^2 - 12x + 5 = 0
4x^2 - 2x - 10x + 5 = 0
2x(2x - 1) - 5 (2x - 1) = 0
(2x - 1) (2x - 5) = 0
2x - 1 = 0 2x - 5 = 0
2x = 1 2x = 5
x = 1/2 x = 5/2
substitute x values into eq2
2x + y = 5
2(1/2) + y = 5
1 + y = 5
y = 4
2x + y = 5
2(5/2) + y = 5
5 + y = 5
y = 0
(x = 1/2,y = 4) and when
(x = 5/2,y = 0)
9)
ReplyDeletex^2+x-12
x^2+4x-3x-12
x(x+4)-3(x+4)
(x+4)(x=-3)
10)
ReplyDelete5x^2+4x+1+(-7x+2)
5x^2+4x+1-7x+2
5x^2-3x+3
13)
ReplyDelete3x^2-48
3(x^2-16)
3(x+4)(x-4)
9)
ReplyDeletex^2+x-12
x^2-3x+4x-12
(x^2-3x)(+4x-12)
x(x-3)+4(x-3)
(x-3)(x+4)
11)
ReplyDelete5x^2+13x-6
5x^2+15x-2x-6
(5x^2+15x)(-2x-6)
5x(x+3)-2(x+3)
(5x-2) (x+3)
#1:
ReplyDeletePerimeter = (L+B) X 2
26 = (L+B)2
26= (x+3)2
13 = x + 3
Therefore length, x, = 10cm
•Area of rectangle: (L x B)
= 10 x 3
= 30cm²
•dA/dx:
2(L+W) =26
L + W = 13
W= 13 –x
Therefore since length = x, and width = (13 – x),
Area= LxB
= (x) (13-x)
= 13x - x²
Therefore dA/dx = 13-2x
•Therefore width of rectangle= 13- 2x
13 = 2x
6.5cm = x
•Therefore maximum area of rectangle = L x B
= 10cm x 6.5cm
= 65cm²
#2:
ReplyDelete3p + 2r = 7……..equation1
p² - 2r = 11…….equation2
using equation1, make r subject: therefore ([7-3p]/2 )= r
p² - 2([7-3p]/2) = 11
p² - (7-3p) = 11
p² - 7 +3p = 11
p² + 3p – 18 = 0
p² - 3p + 6p – 18 =0
after factorizing: (p + 6) (p -3) = 0
therefore values of p= -6 and 3
when p=-6, from equation 1:
3 (-6) +2r =7
2r = 25
r = 12.5
when p = 3, using equation1:
3 (3)+2r = 7
2r = -2
r = -1
#3:
ReplyDelete8x² +3y² = 50…..equation1
2x + y = 5………..equation2
Using equation 2, y = 5-2x
Therefore, 8x² + 3 (5-2x)² = 50
8x² + 75 – 60x +12x² = 50
20x² – 60x +25 = 0……………divide equation by 5.
4x² – 12x + 5 = 0
4x² – 2x – 10x +5 =0
After factorizing, (2x-1) (2x-5) =0
Therefore x = 0.5 and 2.5
Substitute x= 0.5 into equation 2….
2(0.5) + y = 5
1 + y = 5
•Therefore when x=0.5, y=4
Substitute x=2.5 into equation 2…….
2(2.5) +y = 5
5 + y = 5
•Therefore when x= 2.5, y=0.
#4:
ReplyDelete•9x² - 9x +1 in the form a(x + b)² +c:
9x² – 9x +1 = 0
9x² - 9x = -1
x² - [9x/9] = -[1/9]
x² - 1x + 0.25 = - [1/9] + 0.25
(x – 0.5)² = 5/36
Therefore: a=1, b= -0.5, c= -[5/36]
•maximum or minimum value:
Since the value of “a” = 1, which is positive, then the function has minimum value.
•The value at x, which this minimum value occurs = 0.5
#9:
ReplyDeleteX² + x – 12
X² – 3x + 4x – 12
X(x-3) + 4 (x-3)
(x+4) (x-3)
#13:
ReplyDelete3²x -48
3 (x²-16)
3(x-4)(x+4)
#15:
ReplyDeleteF(x) = 3x² -12x +5
3x² -12x = -5
x² - 4x = -5
x² -4x = - 5/3
x² - 4x + 4 = -5/3 +4
(x-2)² = 7/3
In the form a(x+b)² +c : (x-2)² - 7/3, where a= 1, b= -2 and c= -7/3
The function has a minmum value.
Coordinates at which this occurs = (2, -7/3)
Perimeter=2(l+w)
ReplyDelete26=2(L+W)
where W=3CM
SO;
2L + 2(3) = 26
2L=26-6
2L=20
L=10cm
Area of rectangle= b X h
ReplyDelete=10X3
=30CM^2
3p + 2r = 7......(1)
ReplyDeletep^2 - 2r = 11......(2)
From (1),([7-3p]/2 )= r
p^2 - 2([7-3p]/2) = 11
p^2 - (7-3p) = 11
p^2 - 7 +3p = 11
p^2 + 3p – 18 = 0
p^2 - 3p + 6p – 18 =0
(p + 6) (p -3) = 0
p= -6 & p= 3
when p=-6, from (1):
3 (-6) +2r =7
2r = 25
r = 12.5
when p = 3, using(1):
3 (3)+2r = 7
2r = -2
r = -1
3. 8x^2 + 3y^2 = 50......(1)
ReplyDelete2x + y = 5......(2)
From (2)
2x + y = 5
y = 5 - 2x......(3)
sub (3) into (1)
8x^2 + 3(5 - 2x)^2 = 50
8x^2 + 3(25 - 20x + 4x^2) = 50
8x^2 + 75 - 60x + 12x^2 = 50
20x^2 - 60x + 75 - 50 = 0
20x^2 - 60x + 25 = 0
/5: 4x^2 - 12x + 5 = 0
4x^2 - 2x - 10x + 5 = 0
2x(2x - 1) - 5 (2x - 1) = 0
(2x - 1) (2x - 5) = 0
2x - 1 = 0 2x - 5 = 0
2x = 1 2x = 5
x = 1/2 x = 5/2
sub x into (2)
2x + y = 5
2(1/2) + y = 5
1 + y = 5
y = 4
2x + y = 5
2(5/2) + y = 5
5 + y = 5
y = 0
x = 1/2
y = 4
x = 5/2
y = 0
9x² - 9x +1 in the form a(x + b)² +c:
ReplyDelete9x² – 9x +1 = 0
9x² - 9x = -1
x² - [9x/9] = -[1/9]
x² - 1x + 0.25 = - [1/9] + 0.25
(x – 0.5)² = 5/36
a=1, b= -0.5, c= -[5/36]
....maximum or minimum value:
ReplyDeleteSince the value of “a” = 1, which is positive, then the function has minimum value.
....value at x, which this minimum value occurs = 0.5
ReplyDeleteNumber 5.....
ReplyDeletee^2=h^2+f^2-2hfcosx
e^2=(6)^2+(4.2)^2-2(6)(4.2)cos70
e^2=36+17.64-50.4(0.34)
e^2=53.64-17.136
e=squar root of 36.504
e=6.04
the length of HF = 6.04
Number 6.....
ReplyDeleteTriangle 1 (PWF)
tan=opp/adj
.....opp=adj tan tata
opp=28 tan54
opp=28(1.38)
opp=38.64
.....FW=38.64
Triangle 2 (PTF)
ReplyDeletetan=opp/adj
...opp=adj tan tata
opp=28 tan40
opp=28(0.84)
opp=23.52
....FT=23.52
In Conclusion....
ReplyDeleteif.... TW=FW-FT
Then;
TW=38.64-23.52
TW=15.12
Number 9....
ReplyDeletex^2+x-12
x^2+4x-3x-12
x(x+4)-3(x+4)
(x+4)(x=-3)
Number 10....
ReplyDelete5x^2+4x+1+(-7x+2)
5x^2+4x+1-7x+2
5x^2-3x+3
Number 11......
ReplyDelete5x^2+13x-6
5x^2+15x-2x-6
(5x^2+15x)(-2x-6)
5x(x+3)-2(x+3)
(x+3)(5x-2)
Number 1
ReplyDeletegiven that width=3 and perimeter=26, therefore:
P=2L+2W
26=2L+2(3)
26=2L+6
26-6=2L
2L=20
L=20/2
L=10
(ii)
the area:
A=L*W
A=10*3
A=30cm^2
question 9
ReplyDeletex^2+x-12
(x-3)(x+4)
question 10
ReplyDelete(5x^2+4x+1)+(-7x+2)
5x^2+4x+1-7x+2
5x^2+4x-7x+1+2
5x^2-3x+3
question 11
ReplyDelete5x^2+13x-6
(5x-2)(x+3)
question 12
ReplyDelete(8x^4-2x^2)/2x^2
8x^4/2x^2 - 2x^2/2x^2
4x^2-0
4x^2
question 13
ReplyDelete3x^2-48
divide by 3
x^2-16
(x+4)(x-4) difference of two squares
question 16
ReplyDeletey=6x^2+32/x^3
y=6x^2+32x^-3
dy/dx=12x-96x^-4
=12x-96/x^4
question 17
ReplyDeletey=x^2+54/x
y=x^2+54x^-1
dy/dx=2x-54x^-2
=2x-54/x^2
question 18
ReplyDeletey=1+4x^3
dy/dx=12x^2
question 19
ReplyDeletey=x^2+x^3
dy/dx=2x+3x^2
for stationary points
dy/dx=0
3x^2+2x=0
x(3x+2)=0
x=0
3x+2=0
3x=-2
x=-2/3
subt. x=0 and x=-2/3 into y=x^2+x^3
y=(0)^2+(0)^3
y=0
y=(-2/3)^2+(-2/3)^3
=4/9-8/27
=4/27
frist stationary point (0,0)
second stationary point (-2/3,4/27)
Number 5.....
ReplyDeletee^2=h^2+f^2-2hfcosx
e^2=(6)^2+(4.2)^2-2(6)(4.2)cos70
e^2=36+17.64-50.4(0.34)
e^2=53.64-17.136
e=squar root of 36.504
e=6.04
the length of HF = 6.04
question 9
ReplyDeletex^2+x-12
(x-3)(x+4)
2)
ReplyDelete3p + 2r = 7.........eq1
p^2 - 2r = 11.......eq2
eq1 + eq2
3p + p^2 + 2r + - 2r = 18
3p + p^2 = 18
p^2 + 3p - 18 = 0
p^2 + 6p - 3p - 18 = 0
p(p + 6) - 3(p + 6) = 0
(p + 6)(p - 3) = 0
p + 6 = 0 p - 3 = 0
p = -6 p = 3
substitute p values into eq1:
3p + 2r = 7
3(-6) + 2r = 7
-18 + 2r = 7
2r = 25
r = 12.5
3p + 2r = 7
3(3) + 2r = 7
9 + 2r = 7
2r = -2
r = -1
(p= -6,r=12.5)and when(p=3,r= -1)
3. 8x^2 + 3y^2 = 50......(1)
ReplyDelete2x + y = 5......(2)
From (2)
2x + y = 5
y = 5 - 2x......(3)
sub (3) into (1)
8x^2 + 3(5 - 2x)^2 = 50
8x^2 + 3(25 - 20x + 4x^2) = 50
8x^2 + 75 - 60x + 12x^2 = 50
20x^2 - 60x + 75 - 50 = 0
20x^2 - 60x + 25 = 0
/5: 4x^2 - 12x + 5 = 0
4x^2 - 2x - 10x + 5 = 0
2x(2x - 1) - 5 (2x - 1) = 0
(2x - 1) (2x - 5) = 0
2x - 1 = 0 2x - 5 = 0
2x = 1 2x = 5
x = 1/2 x = 5/2
sub x into (2)
2x + y = 5
2(1/2) + y = 5
1 + y = 5
y = 4
2x + y = 5
2(5/2) + y = 5
5 + y = 5
y = 0
x = 1/2
y = 4
x = 5/2
y = 0
question 12
ReplyDelete(8x^4-2x^2)/2x^2
8x^4/2x^2 - 2x^2/2x^2
4x^2-0
4x^2
Number 1 (ii)
ReplyDeletethe area:
A=L*W
A=10*3
A=30cm^2
#19:
ReplyDeletey = x² + 3x³
dy/dx = 2x + 3x²
at stationary points, dy/dx = 0
therefore: 2x + 3x² = 0
therefore: x (2 + 3x) = 0
therefore: either x = 0 or 3x = -2
therefore: x = 0 or x = -2/3
when x= 0, y = 0.
When x = -2/3, y = ?
Y = (-2/3)² + (-2/3)³
Y = 4/27
Therefore coordinates of the stationary points:
(0,0) and (-2/3, 4/27)
This comment has been removed by the author.
ReplyDeletePerimeter=L+L+b+b
ReplyDelete=2L + 2(x)
x=3 -->
26=2L + 2(3)
2L=26 - 6
L=20/2
L=10 cm
L=10cm
ReplyDeleteb= 3cm
Area = L*b
= 10*3
= 30 cm^2
2. 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!!!!!!!
3. 8x^2 + 3y^2 = 50………1
ReplyDelete2x + y = 5………………2
y = 5 – 2x
substitute y = 5 – 2x into equation (1)
8x^2 + 3(5 – 2x)^2 = 50
8x^2 + 3(25 + 4x) = 50
8x^2 + 75 + 12x = 50
Cannot complete
HELP!!!!!!!
4. 9x^2 – 9x + 1
ReplyDeleteIn the form
a (x + b)^2 + c
x = -b/2a
x = -(-9)/18
x = 9/18
x = 2
Therefore (x – 2)^2
Substitute x = 2 into the original equation
To get the value of ‘c’
9(2)^2 – 9(2) + 1 = 19
Write in the form
a (x + b) ^2 + c
9 (x – 2 )^2 + 19
the above equation has a maximum value bescuse the value which 'c' represents is positive
ReplyDeletethe valueof x at which the maximum value is obtained is '+2'
ReplyDelete9.Factorise x^2 + x -12
ReplyDeletecannot complete........
if it could i can't
(2) solve for x and y
ReplyDelete3p+2r=7
p^2-2r=11
4p^3=18
p^3=18/4
p=cube root of 4.5
p= 1.65
3(1.65)+2r=7
2r=7-4.95
2r=2.05
r=2.05/2
r=1.025
perimeter 26cm
ReplyDeletelength x cm
width x equals 3cm
then
2L + 2W = 26
2(x)+ 2(3)=26
2x+6=26
therfore
2x=26-6
2x=20
x=10
area is l*b
area = 10*3=30cm^3
number (1)
ReplyDeleteperimeter = 26
2L + 2W = 26
W=3CM
2L + 2(3) = 26
2L=26-6
2L=20
L=10cm
number (2)
ReplyDelete3p + 2r = 7.........eq1
p^2 - 2r = 11.......eq2
eq1 + eq2
3p + p^2 + 2r + - 2r = 18
3p + p^2 = 18
p^2 + 3p - 18 = 0
p^2 + 6p - 3p - 18 = 0
p(p + 6) - 3(p + 6) = 0
(p + 6)(p - 3) = 0
p + 6 = 0 p - 3 = 0
p = -6 p = 3
substitute p values into eq1:
3p + 2r = 7
3(-6) + 2r = 7
-18 + 2r = 7
2r = 25
r = 12.5
3p + 2r = 7
3(3) + 2r = 7
9 + 2r = 7
2r = -2
r = -1
(p= -6,r=12.5)and when(p=3,r= -1)
number (3)
ReplyDelete8x^2 + 3y^2 = 50......(1)
2x + y = 5......(2)
From (2)
2x + y = 5
y = 5 - 2x......(3)
sub (3) into (1)
8x^2 + 3(5 - 2x)^2 = 50
8x^2 + 3(25 - 20x + 4x^2) = 50
8x^2 + 75 - 60x + 12x^2 = 50
20x^2 - 60x + 75 - 50 = 0
20x^2 - 60x + 25 = 0
/5: 4x^2 - 12x + 5 = 0
4x^2 - 2x - 10x + 5 = 0
2x(2x - 1) - 5 (2x - 1) = 0
(2x - 1) (2x - 5) = 0
2x - 1 = 0 2x - 5 = 0
2x = 1 2x = 5
x = 1/2 x = 5/2
sub x into (2)
2x + y = 5
2(1/2) + y = 5
1 + y = 5
y = 4
2x + y = 5
2(5/2) + y = 5
5 + y = 5
y = 0
x = 1/2
y = 4
x = 5/2
y = 0
9. Factorise
ReplyDeletex^2 + x -12
(x + 4) (x - 3)
10. Evaluate
ReplyDelete(5x^2 + 4x + 1) + (-7x + 2)
multiply the sign outside the bracket
by what is in the bracket
+ (-7x + 2)
= (-7x + 2)
Multiply the first bracket by the second bracket
(5x^2 + 4x + 1) + (-7x + 2)
-35x^3 + 10 x^2 – 28x^2 + 8x – 7x + 2
-35x^3 – 18x^2 + x + 2
i think there is a mistake in question 14 because i dont know wat line is 'AN' can anynoe help me please....
ReplyDelete#8
ReplyDeletey = x^3 - 12 x - 12
dy/dx=3x^2-12
d^2y/dx^2=6x
#8
ReplyDeletedy/dx=3x^2-12
3x^2-12=0
solve for x
perimeter = 26cm
ReplyDeletelength = x
width = 3cm
perimeter = 2(L+w)
= 2(x + 3)
perimeter = 2x + 6
26 = 2x + 6
2x = 26 - 6
2x = 20
x = 20/2
x = 10
length = 10cm
area of rectangle = l x b
ReplyDelete= (2x + 6)(x)
= 2x^2 + 6x
dA/dx = (2)2x^2-1 + 6
dA/dx = 4x + 6
when dA/dx = 0, area is maximum;
0 = 4x + 6
4x = -6
x = -6/4
x = -3/2
when width = -3/2, maximum area is achieved
people, the question did not ask us find the area of the rectangle after finding the length...it is asking us the value of the width which will give us the maximum value for the area of the rectangle...thats when dA/dx = 0
ReplyDeletehey sexy ducklin, when you find the expression for "p", substitute that into the second equation, dont find for "r" one time,you will find it if you put it into the other equation...catch?
ReplyDeletesnowwhite, you have a triangle which is equal to half of the parallelogram. you are right in saying that you dont have a perpendicular height, therefore you cannot use (b x h)/2 but there is another method you can use, just incase it comes in exams: (1/2)(ab sinx), where 'a' and 'b' are two sides of the triangle and 'x' is the angle between them...hopes this helps a bit...
ReplyDeletesexy ducklin, for question 5, use the cosine rule to find HF. you have the two lengths and the angle between them so this is the rule to use...
ReplyDeleteif perimeter = 26
ReplyDelete2L + 2W = 26
BUT W=3CM
2L + 2(3) = 26
2L=26-6
2L=20
L=10cm
Aera = L x W
= 10 x 3
=30cm^2
3p+2r=7
ReplyDeletep^2-2r=11
using elimination method
3p-p^2=-4
-p^2+3p+4=0 *-1
p^2-3p-4=0
p^2-4p+p-4=0
p(p-4)+1(p-4)=0
(p-4)(p+1)=0
p=4 , p=-1
#2
subs.p into 3p+2r=7
p=4
3(4)+2r=7
2r=-5
r=-5/2
p=-1
3(-1)+2r=7
2r=10
r=10/2
r=5
#3:
ReplyDelete8x² +3y² = 50…......eq 1
2x + y = 5..........eq 2
Using eq(2) y = 5-2x
Therefore, 8x² + 3 (5-2x)² = 50
8x² + 75 – 60x +12x² = 50
20x² – 60x +25 = 0..../ eq by 5.
4x² – 12x + 5 = 0
4x² – 2x – 10x +5 =0
After factorizing (2x-1) (2x-5) =0
Therefore x = 0.5 and 2.5
Subst. x= 0.5 into equ 2….
2(0.5) + y = 5
1 + y = 5
Therefore when x=0.5, y=4
Subst x=2.5 into equation 2…….
2(2.5) +y = 5
5 + y = 5
Therefore when x= 2.5, y=0.
hey anybody willing to help and explain question #4 i'm getting trouble with it
ReplyDelete#1: Perimeter = (L+B) X 2
ReplyDelete26 = (L+B)2
26= (x+3)2
13 = x + 3
Therefore length, x, = 10cm
Area of rectangle
ReplyDeleteL=10cm
b= 3cm
therefore:
Area = Lxb
= 10x3
= 30 cm^2
2(L+W) =26
ReplyDeleteL + W = 13
W= 13 –x
Therefore since length = x, and width = (13 – x),
Area= LxB
= (x) (13-x)
= 13x - x²
width of rectangle= 13- 2x
13 = 2x
6.5cm = x
maximum area of rectangle = L x B
= 10cm x 6.5cm
= 65cm²
perimeter = 26cm
ReplyDeletewidth = 3
since perimeter of rectangle = 2L + 2W
26 = 2L + 2(3)
26 = 2L + 6
2L = 26 - 6
2L = 20
L = 20/2
L = 10cm
when L = 10cm
area of a rectangle = L * W
= 10cm * 3cm
= 30cm^2
(14) for number 14 use cosine rule to find the length of HF Hf=6 then they asked for the area of the parallogram and you know that a parallogram has four sides and on right angles so i desided to brake it up into two right angle triangles and a rectangle work out the area for each and add them together the answer i got for the area was 23.7 some please give me some feed back
ReplyDelete9)
ReplyDeletex^2+x-12
x^2-3x+4x-12
(x^2-3x)(+4x-12)
x(x-3)+4(x-3)
(x-3)(x+4)
(8) (5x^2 + 4x + 1) + (-7x + 2)
ReplyDelete5x^2 + 4x + 1 -7x + 2
5x^2 - 3x + 3
10.
ReplyDelete(5x^2+4x+1)+(-7x+2)
5x^2+4x+1-7x+2
5x^2+4x-7x+1+2
5x^2-3x+3
need some help for solving question 19)
ReplyDelete#6 (a)
ReplyDeletefw= tan54 =opp/adj
= tan54 =x/28
=28tan54 = x
=38.5m
#6(b)
ReplyDeletetf=tan40=opp/adj
=tan 40 =x/28
=28tan40=x
=23.4m
therfore
ReplyDeletetw = fw-ft
=38.5-23.4
=15.1m
perimeter= 26 cm.
ReplyDeletelength= x cm.
width= 3cm.
p= (l+w)2
(l+w)2=26.
2x+2w=26.
2x+ 2(3)=26.
2x+6=26.
2x=26-6
2x=20
x=10.
area...(A)=l*w.
A= 10 * 3
A = 30 cm^2.
2(l+w)=26.
l+w=13.
recall that l+x.
w=13-x.
A=l*w
A=(x) (13-x)
A=.....~expand brackets~....
A=13x-x^2
dA/dx= 13-2x.
to find the width, make x the subject of the formula...
w=13-2x.
13=2x
x=13/2
x=6.5cm
maximum area= length * width.
A= 10*6.5
A=65cm^2
13. 3x^2-48.
ReplyDelete3(x^2-16)
#3
ReplyDelete8x^2+3y^2=50..........eq1
2x+y=5................eq2
y=5-2x subt into eq1
8x^2+3(5-2x)=50
8x^2+3(25-20x+4x^2)=50
8x^2+75-60x+12x^2=50
20x^2-60x+75=50
20x^2-60x+75-50=0
20x^2-60x+25=0
4x^2-12x+5=0
4x^2-2x-10x+5=0
2x(2x-1)-5(2x-1)=0
(2x-5)(2x-1)
2x-5=0
2x=5
x=2.5
2x-1=0
2x=1
x=1/2
subx into eq2
2(2.5)+y=5
5+y=5
y=0
2(1/2)+y=5
1+y=5
y=4
therefor
when x=2.5, y= o
and when
x=1/2, y= 4
9. x^2+x-12.
ReplyDeletex^2+4x-3x-12.
x(x+4) -3(x+4).
(x-3) (x+4)
9.
ReplyDeletex^2 + x -12
x^2 - 3x + 4x - 12
x(x - 3) + 4 (x - 3)
(x - 3)(x + 4)
10.
ReplyDelete(5x^2 + 4x + 1) + (-7x + 2)
5x^2 + 4x + 1 -7x + 2
5x^2 + 4x - 7x + 1 + 2
5x^2 - 3x + 3
ques 8
ReplyDeletey = x^3 - 12x - 12
dy/dx = 3x^2 - 12
dy/dx = 0 at stationery pt
3x^2 - 12 = 0
3x^2 -6x + 6x - 12 =0
3x(x - 2)+6(x -2)=0
(x-2)(3x+6)=0
x-2=0 or 3x+6=0
x=2 or x=-2
when x=2 y=-28 (2,-28)
when x=-2 y=4 (-2,4)
12
ReplyDeleteSimplify (8x^4 - 2x^2)/2x^2
4x^2 - 1
11.
ReplyDelete5x^2 + 13x - 6
5x^2 + 15x - 2x - 6
5x(x + 3) - 2(x + 3)
(x + 3)(5x - 2)
12.
ReplyDelete(8x^4 - 2x^2)/2x^2
8x^4/2x^2 - 2x^2/2x^2
4x^4/x^2 - 1
4x^2 - 1
13.
ReplyDelete3x^2 - 48
3[(x^2) - (4)^2]
3[(x + 4)(x - 4)]
5.
ReplyDeletewhen EHF is extracted from the diagram, the shape of the structure is a triangle.
you have an angle(HEF)= 70 degrees.
the side EH (f)= 4.2cm.
the side EF (h)= 6cm.
to find FH, use the cosine rule.
e^2= h^2+f^2 - 2 hf Cos E.
e^2= (6)^2 + (4.2)^2 - 2(6) (4.2) Cos 70.
e^2= 53.64 - 50.4 Cos 70.
e^2= 53.64 - 17.24
e^2= 36.4
square root of e= 6.03 cm.
area.
6.03* 4.2= 25.326 cm^2
can someone help with #20.
ReplyDeletei do not know how to approach this question as i am not understanding how to sketch the graph.
any help would be appreciated.
16.
ReplyDeletey = 6x^2 + 32/x^3
dy/dx = 12x + 96x^2
thank you bornagain 16 that realy helps, can u please solve a ques on vector an post it so i can use it as an example.
ReplyDelete18.
ReplyDeletey = 1 + 4x^3
dy/dx = 12x^2
19.
ReplyDeletey = x^2 + x^3
dy/dx = 2x + 3x^2
dy/dx = 0 for a stationay point
∴ 2x + 3x^2 = 0
x(2 + 3x) = 0
x = 0
2 + 3x = 0
3x = -2
x = -2/3
subet x values into original equation to find coordinates of stationary point
when x = 0
y = x^2 + x^3
y = 0^2 + 0^3
y = 0
when x = -2/3
y = x^2 + x^3
y = (-2/3)^2 + (-2/3)^3
y = 4/27
so corresponding stationary points are
when (x = 0, y = 0)
when (x = -2/3, y = 4/27)
perimeter = 26
ReplyDelete2L + 2W = 26
W=3CM
2L + 2(3) = 26
2L=26-6
2L=20
L=10cm
no. 6
ReplyDeletetan theta= opp/adj
adj*tan theta= opp
using triangle FWP=
28*tan54=38.5
therefore FW= 38.5
using triangle FWT=
28*tan40= 23.5
therefore FT= 23.5
FW-FT= TW
38.5-23.5=15
#1
ReplyDeletep=2L+2W
sub when w=3cm & p=26
p=2x=2(3)
26=2x+6
2x+6=26
2x=26-6
2x=20
x=20/2
i.e L=10cm
#1
ReplyDeleteA=LW
=10*3
=30cm^2
#2
ReplyDelete3p+2r=7......equ 1
p^2-2r=11....equ 2
using subsitution method
3p+2r=7
2r=7-3p
r=7-3p/2
i.e r=7/2-3p/2
#2
ReplyDeletesub into equ 2 when r=7/2-3p/2
p^2-2r=11
p^2-2(7/2-3p/2)=11p^2-7+3p=11
p^2+3p=11+7
3p^3=18
p^3=18/3
p=cube root of 6
p=1.8
#2
ReplyDeletesub into equ 1 when p=1.8
3p+2r=7
3(1.8)+2r=7
5.4+2r=7
2r=7-5.4
r=1.6/2
r=0.8
#3
ReplyDelete8x^2+3y^2=50....equ 1
2x+y=5....equ 2
2x+y=5
y=5-2x
#3
ReplyDeletesub into equ 1 when y=5-2x
8x^2+3(5-2x)^2=50
8x^2+3(25-20x+4x^2)=50
8x^2+75-30x+12x^2=50
8x^2-30x+12x^2=75-50
8x^2+12x^2-30x=25
10x^3=25
x^3=25/10
x=cube root of 2.5
x=1.4
#3
ReplyDeletesub into equ 1 when x=1.4
2x+y=5
2(1.4)+y=5
y=5-2(1.4)
y=5-2.8
y=2.2
#9
ReplyDeletex^2+x-12
x^2+4x-3x-12
x(x+4)-3(x+4)
(x+4)(x-3)
#10
ReplyDelete5x^2+4x+1+(-7x+2)
5x^2+4x+1-7x+2
5x^2+4x-7x+1+2
5x^2-3x+3
2x
#11
ReplyDelete5x^2+13x-6
5x^2+15x-2x-6
5x(x+3)-2(x+3)
(x+3)(5x-2)
#12
ReplyDelete(8x^4-2x^2)/2x^2
8x^4/2x^2 - 2x^2/2x^2
4x^2
#13
ReplyDelete3x^2-48
3x^2/3 - 48/3
x^2-16
(x-4)(x+4)
#14
ReplyDeletei believe something is not right with #14
#16
ReplyDeletey=6x^2+32/x^3
y=6x^2+32x^-3
dy/dx=12x-96x^-4
#17
ReplyDeletey=x^2+54/x
y=x^2+54x^-1
dy/dx=2x-54x^-2
#18
ReplyDeletey=1+4x^3
dy/dx=12x^2
#19
ReplyDeletey=x^2+x^3
dy/dx=2x+3x^2
no. 19
ReplyDeletey=x^2+x^3
dy/dx = 2x+3x^2
for stat pts dy/dx=0
3x^2+2x=0
x(3x+2)=0
x=0
or
3x+2=0
3x=-2
x=-2/3
subt. x=0 and x=-2/3 into y=x^2+x^3
y=(0)^2+(0)^3
y=0
y=(-2/3)^2+(-2/3)^3
=4/9-8/27
=4/27
= (0,0), and (-2/3,4/27)
no9. Factorise x^2 + x -12
ReplyDelete(x - 3)(x + 4)
no11. Factorise 5x^2 + 13x - 6
ReplyDelete(5x + 3) (x + 2)
no13. Factorise 3x^2 - 48
ReplyDelete3(x^2 - 16)
no17. y = x^2 + 54/x, find dy/dx
ReplyDeletedy/dx = 2x -54/x^2
no 18. y = 1 + 4x^3, find dy/dx
ReplyDeletedy/dx = 12x^2
no10. Evaluate (5x^2 + 4x + 1) + (-7x + 2)
ReplyDelete5x^2 + 4x + 1 - 7x + 2
= 5x^2 - 3x + 3
no 8.
ReplyDeletey = x^3 - 12x - 12
dy/dx = 3x^2-12
at stat pts dy/dx= 0
3x^2 - 12 = 0
3x^2= 12
x^2 = 4
x = 2 or -2
sub x= 2 in y
y= (2)^3 - 12(2) - 12
y = - 28
sub x= -2
y= (-2)^3 - 12(-2) -12
y= 4
stat pts = (2,-28) and (-2,4)
5) e^2=h^2+f^2-2hfcosx
ReplyDeletee^2=(6)^2+(4.2)^2-2(6)(4.2)cos70
e^2=36+17.64-50.4(0.34)
e^2=53.64-17.136
e=squar root of 36.504
e=6.04
the length of HF = 6.04
Area of rectangle
ReplyDeleteL=10cm
b= 3cm
therefore:
Area = Lxb
= 10x3
= 30 cm^2
7)Triangle (ABD)
ReplyDeletetan tayta=opp/adj
i.e opp=adj tan tayta
opp=2.5 tan75
opp=2.5(3.73)
opp=8.4
AB=8.4
1)
ReplyDeletewidth (W)=3
perimeter (p)=26
P=2L+2W
26=2L+2(3)
26=2L+6
26-6=2L
2L=20
L=20/2
L=10
Number 11......
ReplyDelete5x^2+13x-6
5x^2+15x-2x-6
(5x^2+15x)(-2x-6)
5x(x+3)-2(x+3)
(x+3)(5x-2)
if perimeter = 26
ReplyDelete2L + 2W = 26
BUT W=3CM
2L + 2(3) = 26
2L=26-6
2L=20
L=10cm
Aera = L x W
= 10 x 3
=30cm^2
2)
ReplyDelete3p + 2r = 7.........eq1
p^2 - 2r = 11.......eq2
eq1 + eq2
3p + p^2 + 2r + - 2r = 18
3p + p^2 = 18
p^2 + 3p - 18 = 0
p^2 + 6p - 3p - 18 = 0
p(p + 6) - 3(p + 6) = 0
(p + 6)(p - 3) = 0
p + 6 = 0 p - 3 = 0
p = -6 p = 3
substitute p values into eq1:
3p + 2r = 7
3(-6) + 2r = 7
-18 + 2r = 7
2r = 25
r = 12.5
3p + 2r = 7
3(3) + 2r = 7
9 + 2r = 7
2r = -2
r = -1
d^2y/dx^2=3x^2-12
ReplyDeleteusing loop
d^2y/dx^2=6x
d^2y/dx^2=6/x
sub when x=2
d^2y/dx^2=6/2
=3
sub when x=-2
d^2y/dx^2=6/-2
=-3
2) 3p+2r=7
ReplyDeletep^2-2r=11
using method of elimination
3p-p^2=-4
-p^2+3p+4=0 *-1
p^2-3p-4=0
p^2-4p+p-4=0
p(p-4)+1(p-4)=0
(p-4)(p+1)=0
p=4 , p=-1
sub p into 3p+2r=7
p=4
3(4)+2r=7
2r=-5
r=-5/2
p=-1
3(-1)+2r=7
2r=10
r=5
11.
ReplyDelete5x^2+13x-6
5x^2+15x-2x-6
(5x^2+15x)(-2x-6)
5x(x+3)-2(x+3)
(x+3)(5x-2)
8) (5x^2 + 4x + 1) + (-7x + 2)
ReplyDelete5x^2 + 4x + 1 -7x + 2
5x^2 - 3x + 3
12)
ReplyDelete(8x^4-2x^2)/2x^2
8x^4/2x^2 - 2x^2/2x^2
4x^2-0
4x^2
9)
ReplyDeletex^2+x-12
x^2-3x+4x-12
(x^2-3x)(+4x-12)
x(x-3)+4(x-3)
(x-3)(x+4)