- f(x) = 2 - 3x -2x^2 bring to the form a(x + b)^2 + c and hence determine the coordinates of the turning point. Is the t.p. a max or min justify your answer.
- Solve the equations 3x +y = 14 and 2x^2 -xy = 3
- Express 1 - 4x - 2x^2 in the form a - b(x + c) hence when ix the function a maximum
- CDB is a right angle triangle, BC is 5 metres and BCD is 40 degrees and BDC is 90 degrees. Calculate the length of BD and DC. Show that the area of the triangle is 12.5sin 40cos40.
- Solve the equation 5y^2 = 8y - 2
- MLO is a triangle with MNL a right angle triangle and NLO another right angle triangle. ML is 26 cm, NL is 10 cm, MOL is 35 degrees. Calculate the length of MN and MO.
- Solve the simultaneous equations 2x^2 + y^2 = 33 and x + y = 3
- Solve the simultaneous equations x + 1 = 2y and x^2 - 3y = 4
- EFGH is a parallelogram EFI is 40 degrees, EF = 8m, EI is a perpendicular line to FG such that IG = 5m. Calculate the length FI, EI and the area of EFGH.
- ABCD is a trapezium. AB = 12m, AD = 1.5m, BC = 3m and AD is parallel to DC. I is a point on BC such that CID is 90 degrees. Calculate the angle CDI and the length of DC.
Sunday, November 29, 2009
Revision questions
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5) 5y^2=8y-2
ReplyDelete5y^2-8y+2=0
from root equation -b+/- the root of b squared -
4ac devided by 2a
=8+- root(64-40)devided by 10
=8 +-root of 24/10
root of 24/10=1.55
roots are 6.45 and 9.55
6)
ReplyDeleteTAKING TRI NLO
sin 35=NL/10
10*sin35=nl
NL=5.74
takin tri MLN
NL^2 +MN^2=26^2
mn^2=26^2 -5.74^2
mn^2=640.05
MN=25.29
MO = MN + N0
= 31.04
30
ReplyDelete1-4x-2x^2 in the form a-b(x+c)
=1-2x (x+2)
where the max pt is at x= -2
from (5) 5y^2 = 8y - 2
ReplyDelete5y^2-8y = -2
y(5y-8)= -2
y= -2/5y-8
y^2= -2/5-8
y^2 = -2/-3
y^2 = 2/3
y = 2/3^1/2
y = 0.816
7) 2x^2 + y^2 = 33 .....(1)
ReplyDeletex + y = 3.....(2)
from (2) y = 3-x
substitute y = 3-x in (1)
which would be; 2x^2 + (3-x)^2 = 33
2x^2 + 9 -6x + x^2 = 33
3x^2 -6x + 9 = 33
3x^2 - 6x = 24
3x^2 - 6x - 24 = 0
3x^2 +6x -12x -24 =0
3x(x +2) -12 (x +2) =0
3x -12=0
x+2 =0
x= -2
x=4
substitute x=4 and x= -2 in y= 3-x
y=-1
and y = 5
from 4) CDB is a right angle triangle, BC is 5 metres and BCD is 40 degrees and BDC is 90 degrees. Calculate the length of BD and DC. Show that the area of the triangle is 12.5sin 40cos40.
ReplyDeletesin angle = opp/hyp
sin 40 = BD/5
BD = sin 40 * 5
BD = 3.214 metres
4)
ReplyDeletecos angle = adj/hyp
cos 40 = DC / 5
DC = cos 40 * 5
DC =3.83
in question 4) 12.5sin 40cos40 = area of triangle
ReplyDeletearea of triangle = 1/2 (base * height)
1/2 (BD *DC) = 6.15 m^2
12.5sin 40cos40 is also equal to 6.15 m^2
from (8)
ReplyDeletex + 1 = 2y ....(1)
x^2 - 3y = 4...(2)
from (1) y = x+1/2
sub. y = x+1/2 in (2)
x^2 - 3(x+1/2) =4
2x^2 - 3x -3 =4
2x^2 -3x -3-4=0
2x^2 -3x -7=0
using quadratic formula ;= -b +/- ((b^2 - 4 a*c)^1/2 )/2a
x=8.3 and -1.1375
sub. x=8.3 and -1.1375 in (1)
y = -0.06875 and 4.65
2) 3x +y = 14....(1)
ReplyDelete2x^2 -xy = 3..(2)
from (1) y = 14-3x
substitute y = 14-3x in (2)
2x^2 -14x + 3x^2 = 3
5x^2 -14x -3 = 0
5x^2 -15x +x -3 = 0
5x(x-3) 1(x-3) = 0
5x +1 =0
x-3 =0
x= 3 and -1/5
sub. x= 3 and -1/5 in (1)
y = 5 and 14.6
for ques 7
ReplyDelete2x^2 + y^2 = 33 eq1
x + y = 3 eq 2
from eq2 y = 3-x
sub y = 3-x in eq1
2x^2 + (3-x)^2 = 33
2x^2 + 9 -6x + x^2 = 33
3x^2 -6x + 9 = 33
3x^2 - 6x = 24
3x^2 - 6x - 24 = 0
3x^2 +6x -12x -24 =0
3x(x +2) -12 (x +2) =0
3x -12=0
x+2 =0
x= -2
x=4
sub x=4 and x= -2 in y= 3-x
y= -1
and y= 5
5)
ReplyDelete5y^2=8y-2
5y^2-8y+2=0
4ac devided by 2a
=8+- root(64-40)devided by 10
=8 +-root of 24/10
root of 24/10=1.55
roots are 6.45 and 9.55
question 2
ReplyDelete3x + y=14...eq(1)
2x^2 - xy=3...eq(2)
from eq(1)
3x + y=14
y=14-3x...eq(3)
substitute eq(3) into eq(2)
2x^2 - x(14-3x)=3
2x^2-14x+3x^2=3
5x^2-14x-3=0
(5x+1)(x-3)=0
5x+1=0
5x=-1
x=-1/5
x-3=0
x=3
substitute x=-1/5 and x=3 into eq(1)
3(-1/5)+y=14
-3/5+y=14
multiply throughout by 5
-3+5y=70
5y=70+3
5y=73
y=73/5
3(3)+y=14
9+y=14
y=14-9
y=5
when x=-1/5, y=73/5
when x=3, y=5
no: 5
ReplyDelete5y^2-8y = -2
y(5y-8)= -2
y= -2/5y-8
y^2= -2/5-8
y^2 = -2/-3
y^2 = 2/3
y = 2/3^1/2
y = 0.816
from question 4;
ReplyDeletecos angle = adj/hyp
cos 40 = DC / 5
DC = cos 40 * 5
DC =3.83
Solve the equations 3x +y = 14 and 2x^2 -xy = 3
ReplyDelete(2)3x+ y=14 (1)
2x^2 -xy=3 (2)
from (1) y= 14- 3x (3)
sub (3) into (2)
2x^2-x(14- 3x)=3
2x^2- 14x+ 3x^2=3
5x^2- 14x- 3=0
5x^2 - 15x +1x -3=0
5x(x- 3)+1 (x- 3)=0
(5x +1) (x- 3)=0
5x +1=0 x-3=0
5x= -1 x=3
x= -1/5
sub x=-1/5 into (3) sub x=3
y= 14- 3(-1/5) y=14- 3(3)
y= 14+ 3/5 y=14- 9
y= 83/5 y=5
therefore when x= -1/5, y= 83/5
when x=3, y= 5
from no.1
ReplyDeletef(x)= 2x^2-3x-2
a b c
h=b/2a... -3/2(2)= -3/4
h= 4ac-b^2/2a.... 4(2)(-2)-(-3^2)/4(2)= -7/4
a(x+h)^2 +k
2(x - 3/4)^2 - 7/9
max pt
1
ReplyDeletef(x) = 2 - 3x - 2x^2
-2( x^2 + (3/2)x + (3/4)^2 ) + 2 + 2(3/4)^2
-2( x + (3/4))^2 + 3.125
Maximum turning point at(-(9/16) , 3.125)
ReplyDeleteIt is max because 3.125 is positive
You can also know the nature of the turning point before completing the square . By looking at the coefficient of x^2 . If it is positive then it is min and if it is negative max.
ReplyDeleteQuestion 9
ReplyDeleteSOLVING SIMULTANEOUS EQUATIONS (one linear and one quadratic) STEP BY STEP By REAPER
STEP 1
NAME THE EQUATIONS BY NUMBERING THEM
x + 1 = 2y (1)(this eqn is now called eqn 1 )x^2 - 3y = 4 (2)(this eqn is now called eqn 2)
QUESTION 9
ReplyDeletex + 1 = 2y (1)
x^2 -3y = 4 (2)
STEP 2
Transpose the linear equation to make either x or y the subject and call the resulting eqn,eqn 3
From eqn 1
x = 2y - 1 (3)
(7)2x^2 + y^2=33 (1)
ReplyDeletex+ y= 3 (2)
from (1) y= 3-x (3)
sub (3) into (1)
2x^2 + (3- x)^2= 33
2x^2 + (3- x)(3- x)= 33
2x^2 + 9- 6x + x^2= 33
3x^2- 6x +9= 33
3x^2- 6x - 24=0
3x^2- 6x +12x- 24=0
3x(x- 2) +12(x- 2)=0
(3x+ 12)(x- 2)=0
3x+ 12=0 x- 2=0
3x= -12 x= 2
x= -12/3
x= -4
sub x= -4 into (3) sub x=2 into (3)
y=3 - (-4) y= 3 -2
y=7 y=1
question 9
ReplyDeleteSTEP 3
substitute eqn 3 into the quadratic eqn
sub x = 2y - 1 into x^2 - 3y = 4
(2y - 1)( 2y -1) - 3y = 4
STEP 4
ReplyDeleteSimplify the eqn
4y^2 - 4y + 1 - 3y = 4
4y^2 - 7y - 3 = 0
QUESTION 9
ReplyDeleteSTEP 5
Solve for y by either factorisation or quadratic formula
4y^2 - 7y - 3 = 0 cannot be factorised
The quadratic formula gives
x = - 7.4 and x = -4.8
question 9
ReplyDeleteI made a mistake in step 6 it supposed to be
y = -7.4 and y = -4.8
STEP 6
ReplyDeleteSubstitute the y values in eqn 3 and get corresponding x values
sub y = -7.4 in eqn 3
x = -15.8
sub y = -4.8 in eqn 3
x = -10.6
STEP 7 (FINAL STEP)
ReplyDeleteWrite answer in this format
when y = -7.4 ,x = -15.8 and when y = -4.8, x = -10.6
7)
ReplyDelete2x^2 + y^2 = 33 .....eq(1)
x + y = 3............eq(2)
from (2) y = 3-x
substitute y = 3-x in eq(1)
2x^2 + (3-x)^2 = 33
2x^2 + 9 -6x + x^2 = 33
3x^2 -6x + 9 = 33
3x^2 - 6x = 24
3x^2 - 6x - 24 = 0
3x^2 +6x -12x -24 =0
3x(x +2) -12 (x +2) =0
(3x+ 12)(x- 2)=0
3x+ 12=0 x- 2=0
3x= -12 x= 2
x= -12/3
x= -4
sub x= -4 into (1) sub x=2 into (1)
y=3 - (-4) y= 3 -2
y=7 y=1
2)
ReplyDelete3x+y=14........(1)
2x^2-xy=3......(2)
frm eqn (1) y=14-3x sub into 2
2x^2-x(14-3x)=3
2x^2-14x+3x^2=3
5x^2-14x-3=0
(5x+1)(x-3)=0
x=-1/5
or x=3
when x=-1/5 sub into eq'n (1)
3(1/5)+y=14
(x by 5) -3+5y=70
5y=73
y=73/5
and x+3
3(3)+y=14
y=5
5)
ReplyDelete5y^2=8y-2
5y^2-8y=2=0
using y= (-b+-√b^2-4ac)/2a
y= (8+-√8^2-4(5)(2))/2(5)
either y=(8+√64+40)/10
y= (8+10.2)/10=1.82
or y= (8-10.2)/10=-0.22
8)
ReplyDeletex+1=2y.....(1)
x^2-3y=4...(2)
from eq'n (1) x=2y-1 sub into (2)
(2y-1)^2-3y=4
4y^2-4y+1-3y=4
4y^2-7y-3=0
(4y-3)(y-1)
either y=3/4
or y=1
when y=3/4 sub into (1)
x+1=2(3/4)
x+1=3/2
x=2.5
when y=1
x+1=2(1)
x=1
(8)
ReplyDeletex + 1 = 2y ....(1)
x^2 - 3y = 4...(2)
from (1) y = x+1/2
sub. y = x+1/2 in (2)
x^2 - 3(x+1/2) =4
2x^2 - 3x -3 =4
2x^2 -3x -3-4=0
2x^2 -3x -7=0
-b +/- ((b^2 - 4 a*c)^1/2 )/2a
x=8.3 and -1.1375
sub. x=8.3 and -1.1375 in (1)
y = -0.06875 and 4.65
3x +y = 14 (1)
ReplyDelete2x^2 -xy = 3 (2)
from equation 1:
y= 14-3x (3)
substitute equation 3 into equation 2:
2x^2 -x(14-3x) = 3
(2) 2x^2-14x+3x^2 = 3
2x^2 +3x^2 -14x -3 = 0
factorise the equation:
5x^2 -14x -3 =0
5x^2 +15x -1x -3 =0
5x(x+3) -1(x+3)=0
(x+3)(5x-1) = 0
find the x values:
x+3 =0
x = -3
5x-1 =0
5x=1
x=1/5
find the y values:
when x is -3, y is:
3(-3) +y = 14
-9+y =14
y = 14+9
y =23
when x is 1/5, y is:
3(1/5) +y = 14
3/5 +y = 14
y= 14 - 3/5
y= 13 2/5
SO therefor the points are (-3,23) and (1/5, 13 2/5).
2x^2 + y^2 = 33 (1)
ReplyDeletex + y = 3 (2)
from equation 2:
y = 3-x (3)
substitute equation 3 into equation 1:
2x^2 + (3-x)^2 = 33
2x^2 + x^2 -6x +9 =33
3x^2 -6x +9-33 =0
3x^2 -6x -24 =0
factorise the equation:
3x^2 -6x -24 =0
3x^2 -12x +6x -24 =0
3x(x-4) +6(x-4) =0
(3x+6) (x-4) =0
find the x values:
3x+6 =0
3x = -6
x = -2
x-4 = 0
x=4
find the y values:
when x is -2, y is:
-2 + y = 3
y = 3+2
y = = 5
when x is 4, y is:
4 + y =3
y= 3-4
y= -1
SO therefore the points are (-2,5) and (4,-1)
Solve the equation 5y^2 = 8y - 2
ReplyDelete(5)5y^2 -8y + 2=0
you would have to use the quadratic formula because the trial and error method is not working therefore,
y= (-b +/- √b^2 -4ac)/2a
y= (+8 +/- √(-8)^2 - 4(5)(2))/10
y= (8 +/- √64 - 40)/10
y= (8 +/- √24)/10
y= (8 + √24)/10
y=1.28
y= (8- √24)/10
y=0.31
therefore y= 1.28, y= 0.31
2)
ReplyDelete3x +y = 14....(1)
2x^2 -xy = 3..(2)
from (1) y = 14-3x
substitute y = 14-3x in (2)
2x^2 -14x + 3x^2 = 3
5x^2 -14x -3 = 0
5x^2 -15x +x -3 = 0
5x(x-3) 1(x-3) = 0
5x +1 =0
x-3 =0
x= 3 and -1/5
sub. x= 3 and -1/5 in (1)
y = 5 and 14.6
QUESTION 2:
ReplyDelete3x +y = 14.....eq1
2x^2 -xy = 3.....eq2
from eq1 y = 14-3x
substitute y = 14-3x in eq2
2x^2- x(14-3x)=3
2x^2-14x + 3x^2 = 3
5x^2-14x -3 = 0
5x^2-15x +x -3 = 0
5x(x-3)+1(x-3) = 0
(5x +1) (x-3)=0
5x+1=0 x=3
x=-1/5
x-1/5sub. x= 3 and -1/5 in eq1
y = 5 and 14.6
7)
ReplyDelete2x^2 + y^2 = 33 .....1
x + y = 3.....2
from 2... y = 3-x
substitute y = 3-x in 1
which would be; 2x^2 + (3-x)^2 = 33
2x^2 + 9 -6x + x^2 = 33
3x^2 -6x + 9 = 33
3x^2 - 6x = 24
3x^2 - 6x - 24 = 0
3x^2 +6x -12x -24 =0
3x(x +2) -12 (x +2) =0
3x -12=0
x+2 =0
x= -2
x=4
substitute x=4 and x= -2 in y= 3-x
y=-1
and y = 5
2.
ReplyDelete3x+y=14........(1)
2x^2-xy=3......(2)
frm eqn (1) y=14-3x sub into 2
2x^2-x(14-3x)=3
2x^2-14x+3x^2=3
5x^2-14x-3=0
(5x+1)(x-3)=0
x=-1/5
or x=3
when x=-1/5 sub into eq'n (1)
3(1/5)+y=14
(x by 5) -3+5y=70
5y=73
y=73/5
and x+3
3(3)+y=14
y=5
1
ReplyDeletef(x) = 2 - 3x - 2x^2
-2( x^2 + (3/2)x + (3/4)^2 ) + 2 + 2(3/4)^2
-2( x + (3/4))^2 + 3.125
Maximum turning point at(-(9/16) , 3.125)
ReplyDeleteIt is max because 3.125 is positive
You can also know the nature of the turning point before completing the square . By looking at the coefficient of x^2 . If it is positive then it is min and if it is negative max.
ReplyDelete5) 5y^2=8y-2
ReplyDelete5y^2-8y+2=0
from root equation -b+/- the root of b squared -
4ac devided by 2a
=8+- root(64-40)devided by 10
=8 +-root of 24/10
root of 24/10=1.55
roots are 6.45 and 9.55
7) 2x^2 + y^2 = 33 .....(1)
ReplyDeletex + y = 3.....(2)
from (2) y = 3-x
substitute y = 3-x in (1)
which would be; 2x^2 + (3-x)^2 = 33
2x^2 + 9 -6x + x^2 = 33
3x^2 -6x + 9 = 33
3x^2 - 6x = 24
3x^2 - 6x - 24 = 0
3x^2 +6x -12x -24 =0
3x(x +2) -12 (x +2) =0
3x -12=0
x+2 =0
x= -2
x=4
substitute x=4 and x= -2 in y= 3-x
y=-1
and y = 5
question 2
ReplyDelete3x + y=14...eq(1)
2x^2 - xy=3...eq(2)
from eq(1)
3x + y=14
y=14-3x...eq(3)
substitute eq(3) into eq(2)
2x^2 - x(14-3x)=3
2x^2-14x+3x^2=3
5x^2-14x-3=0
(5x+1)(x-3)=0
5x+1=0
5x=-1
x=-1/5
x-3=0
x=3
substitute x=-1/5 and x=3 into eq(1)
3(-1/5)+y=14
-3/5+y=14
multiply throughout by 5
-3+5y=70
5y=70+3
5y=73
y=73/5
3(3)+y=14
9+y=14
y=14-9
y=5
when x=-1/5, y=73/5
when x=3, y=5
question 1
ReplyDeletef(x) = 2 - 3x - 2x^2
-2( x^2 + (3/2)x + (3/4)^2 ) + 2 + 2(3/4)^2
-2( x + (3/4))^2 + 3.125
Maximum turning point at(-(9/16) , 3.125)
becuz 3.125 is positive
question 2
ReplyDelete3x+y=14........(1)
2x^2-xy=3......(2)
from equation (1) y=14-3x sub into 2
2x^2-x(14-3x)=3
2x^2-14x+3x^2=3
5x^2-14x-3=0
(5x+1)(x-3)=0
x=-1/5
or x=3
when x=-1/5 sub into equation (1)
3(1/5)+y=14
(x by 5) -3+5y=70
5y=73
y=73/5
and x+3
3(3)+y=14
y=5
5) 5y^2=8y-2
ReplyDelete5y^2-8y+2=0
from root equation -b+/- the root of b squared -
4ac devided by 2a
=8+- root(64-40)devided by 10
=8 +-root of 24/10
root of 24/10=1.55
roots are 6.45 and 9.55
7) 2x^2 + y^2 = 33 .....(1)
ReplyDeletex + y = 3.....(2)
from (2) y = 3-x
substitute y = 3-x in (1)
which would be; 2x^2 + (3-x)^2 = 33
2x^2 + 9 -6x + x^2 = 33
3x^2 -6x + 9 = 33
3x^2 - 6x = 24
3x^2 - 6x - 24 = 0
3x^2 +6x -12x -24 =0
3x(x +2) -12 (x +2) =0
3x -12=0
x+2 =0
x= -2
x=4
substitute x=4 and x= -2 in y= 3-x
y=-1
and y = 5
2) 3x +y = 14....(1)
ReplyDelete2x^2 -xy = 3..(2)
from (1) y = 14-3x
substitute y = 14-3x in (2)
2x^2 -14x + 3x^2 = 3
5x^2 -14x -3 = 0
5x^2 -15x +x -3 = 0
5x(x-3) 1(x-3) = 0
5x +1 =0
x-3 =0
x= 3 and -1/5
sub. x= 3 and -1/5 in (1)
y = 5 and 14.6
7) 2x^2 + y^2 = 33 .....(1)
ReplyDeletex + y = 3.....(2)
from (2) y = 3-x
substitute y = 3-x in (1)
which would be; 2x^2 + (3-x)^2 = 33
2x^2 + 9 -6x + x^2 = 33
3x^2 -6x + 9 = 33
3x^2 - 6x = 24
3x^2 - 6x - 24 = 0
3x^2 +6x -12x -24 =0
3x(x +2) -12 (x +2) =0
3x -12=0
x+2 =0
x= -2
x=4
substitute x=4 and x= -2 in y= 3-x
y=-1
and y = 5
8)
ReplyDeletex+1=2y.....(1)
x^2-3y=4...(2)
from eq'n (1) x=2y-1 sub into (2)
(2y-1)^2-3y=4
4y^2-4y+1-3y=4
4y^2-7y-3=0
(4y-3)(y-1)
either y=3/4
or y=1
when y=3/4 sub into (1)
x+1=2(3/4)
x+1=3/2
x=2.5
when y=1
x+1=2(1)
x=1
number 7
ReplyDelete2x^2 + y^2 = 33 .....(1)
x + y = 3.....(2)
from (2) y = 3-x
substitute y = 3-x in (1)
which would be; 2x^2 + (3-x)^2 = 33
2x^2 + 9 -6x + x^2 = 33
3x^2 -6x + 9 = 33
3x^2 - 6x = 24
3x^2 - 6x - 24 = 0
3x^2 +6x -12x -24 =0
3x(x +2) -12 (x +2) =0
3x -12=0
x+2 =0
x= -2
x=4
substitute x=4 and x= -2 in y= 3-x
y=-1
and y = 5
number (2)
ReplyDelete3x +y = 14....(1)
2x^2 -xy = 3..(2)
from (1) y = 14-3x
substitute y = 14-3x in (2)
2x^2 -14x + 3x^2 = 3
5x^2 -14x -3 = 0
5x^2 -15x +x -3 = 0
5x(x-3) 1(x-3) = 0
5x +1 =0
x-3 =0
x= 3 and -1/5
sub. x= 3 and -1/5 in (1)
y = 5 and 14.6
number (5)
ReplyDelete5y^2=8y-2
5y^2-8y+2=0
from root equation -b+/- the root of b squared -
4ac devided by 2a
=8+- root(64-40)devided by 10
=8 +-root of 24/10
root of 24/10=1.55
roots are 6.45 and 9.55
number (8)
ReplyDeletex + 1 = 2y ....(1)
x^2 - 3y = 4...(2)
from (1) y = x+1/2
sub. y = x+1/2 in (2)
x^2 - 3(x+1/2) =4
2x^2 - 3x -3 =4
2x^2 -3x -3-4=0
2x^2 -3x -7=0
-b +/- ((b^2 - 4 a*c)^1/2 )/2a
x=8.3 and -1.1375
sub. x=8.3 and -1.1375 in (1)
y = -0.06875 and 4.65
number (6)
ReplyDeleteTAKING TRI NLO
sin 35=NL/10
10*sin35=nl
NL=5.74
takin tri MLN
NL^2 +MN^2=26^2
mn^2=26^2 -5.74^2
mn^2=640.05
MN=25.29
MO = MN + N0
= 31.04
#2
ReplyDeleteSolve the equations 3x +y = 14 and 2x^2 -xy = 3
3x+ y=14 (1)
2x^2 -xy=3 (2)
from (1) y= 14- 3x (3)
sub (3) into (2)
2x^2-x(14- 3x)=3
2x^2- 14x+ 3x^2=3
5x^2- 14x- 3=0
5x^2 - 15x +1x -3=0
5x(x- 3)+1 (x- 3)=0
(5x +1) (x- 3)=0
5x +1=0 x-3=0
5x= -1 x=3
x= -1/5
sub x=-1/5 into (3) sub x=3
ReplyDeletey= 14- 3(-1/5) y=14- 3(3)
y= 14+ 3/5 y=14- 9
y= 83/5 y=5
when x= -1/5, y= 83/5
when x=3, y= 5
1). f(x)=2-3x-2x^2
ReplyDelete-2x^2-3x+2
-2(x^2+3/2x-1)
-2(x+3/4)^2-1
t.p=(x+3/4)=0
x= -3/4
(x+3/4)(x+3/4)
x^2+3/2x+9/16
-2(x+3/4)^2-25/16
-2(x+3/4)^2-2(25/16)
-2(x+3/4)^2+25/8
the turning pts. (-3/4,25/8)
dy/dx= -3-4x
d2y/dx2= -4
it has a max. pt. because d2ydx2<0
2.
ReplyDelete3x + y = 14_____1
2x^2 - xy = 3___2
1: 3x + y = 14
y = 14 - 3x_____3
subst 3 into 2:
2x^2 - x(14 - 3x) = 3
2x^2 -14x + 3x^2 = 3
5x^2 - 14x - 3 = 0
5x^2 -15x + x - 3 = 0
5x(x - 3) + 1(x - 3) = 0
(x - 3)(5x + 1) = 0
x - 3 = 0
x = 3
5x + 1 = 0
5x = -1
x = -1/5
xubst x values into eq3:
when x = 3
y = 14 - 3x
y = 14 - 3(3)
y = 5
when x = -1/5
y = 14 - 3x
y = 14 - 3(-1/5)
y = 14.6
so corresponding values are
when (x = 3, y = 5)
when (x = -1/5, y = 14.6)
7.
ReplyDelete2x^2 + y^2 = 33_____1
x + y = 3___________2
2: y = 3 - x________3
subst 3 into 1:
2x^2 + (3 - x)^2 = 33
2x^2 + 9 - 6x + x^2 = 33
3x^2 - 6x + 9 - 33 = 0
3x^2 - 6x - 24 = 0
/3: x^2 - 2x -8 = 0
x^2 + 2x - 4x - 8 = 0
x(x + 2) - 4 (x + 2) = 0
(x + 2)(x - 4) = 0
x + 2 = 0
x = -2
x - 4 = 0
x = 4
subst x values into eqn3
when x = -2
y = 3 - x
y = 3 - - 2
y = 5
when x = 4
y = 3 - x
y = 3 - 4
y = -1
so corresponding values are:
when (x = -2,y = 5)
when (x = 4,y = -1)
#2.
ReplyDelete3x+y=14.....eq1
2x^2-xy=3...eq2
from eq1 y=14-3x.....eq3
subst. eq3 into eq2
2x^2-x(14-3x)=3
2x^2-14x+3x^2=3
5x^2-14x-3=0
(5x+1)(x-3)
5x+1=0
5x=-1
x=-1/5
x-3=0
x=3
when x=-1/5
3(-1/5)+y=14
-3/5+y=14
multiply across by 5
-3+5y=70
5y=70+3
5y=73
y=73/5
when x=3
3(3)+y=14
9+y=14
y=14-9
y=5
(-1/5,73/5) (3,5)
#4
ReplyDeletesin 40=BD/5
BD=sin40*5=3.214m
cos 40=DC/5
DC=cos40*5=3.83
12.5 sin40 cos40=Area
area of tri. = 1/2 (b*h)
=1/2 (3.214*3.83)=6.15m^2
12.5 sin40 cos40=6.15m^2
#5.
ReplyDelete5y^2-8y+2=0
a=5 b=-8 c=2
y=[-b+/-root b^2-4ac]/2a
=[8+/- root (-8)^2-4(5)(2)]/2(5)
=[8+/- root (64-40)]/10
=[8+/- root 24]/10
[8+ root 24]/10=1.28
[8- root 24]/10=0.31
#6.
ReplyDeletesin35=NL/10
10*sin35=NL
NL=5.74
NL^2+MN^2=26^2
MN^2=26^2-5.74^2
MN^2=640.05
MN=25.29
MO=MN+NO=31.04
#7.
ReplyDelete2x^2+y=33....eq1
x+y=3........eq2
from eq2....y=3-x....eq3
subst. eq3 into eq1
2x^2+(3-x)^2=33
2x^2+(3-x)(3-x)=33
2x^2+x^2+9-6x=33
3x^2-12x+6x-24=0
3x(x-4)+6(x-4)
(3x+6)(x-4)
3x+6=0
3x=-6
x=-2
x-4=0
x=4
when x=-2
-2+y=3
y=5
when x=4
4+y=3
y=-1
(-2,5) (4,-1)
This comment has been removed by the author.
ReplyDelete#8.
ReplyDeletex+1=2y.....eq1
x^2-3y=4...eq2
from eq1....x=2y-1....eq3
subst. eq3 into eq2
(2y-1)^2-3y=4
(2y-1)(2y-1)-3y=4
4y^2-4y+1-3y=4
4y^2-7y-3=0
a=4 b=-7 c=-3
y=[-b+/- root b^2-4ac]/2a
=[7+/- root 49-48]/8
=[7+/- root 97]/8
[7+ root 97]/8=2.12
[7- root 97]/8=-0.36
when y=2.12
x+1=2(2.12)
x=4.24-1=3.24
when y=-0.36
x+1=2(-0.36)
x=-0.72-1=-1.72
(3.24,2.12) (-1.72,-0.36)
#4
ReplyDeletesinx=opp/hyp
opp=5 sin40
opp=5(0.64)
BD=3.2m
#4
ReplyDeleteh^2=o^2+a^2
o^2+a^2=h^2
a^2=h^2-o^2
a^2=5^2-3.2^2
a^2=25-10.24
a^2=14.76
a=3.8m
i.e DC=3.8m
#5
ReplyDelete5y^2=8y-2
5y62-8Y=-2
Y=-2/3