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LHS

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Angle=

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sqrt(3x+4)=2+ sqrt(x)

square






rearrange


square



/4x
x=4

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LHS =

Take LCD


















LHS = RHS

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D.x^2-25
(x+5)(x-5)=0
so x=5 or x=-5

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slope of the given line = 2/5
passes through (0,0)
y-0=2/5(x-0)
y=(2/5)x
-(2/5)x+y=0

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x y z are in geometric
x/y = y/z
OR y^2=xz

x^2, y^2, z^2
x^2, xz, z^2
If they are in geometric sequence then
xz/x^2 = z^2/xz
z/x = z/x
The ratios are common
so x^2,y^2 & z^2 is a GP

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x1 y1 x2 y2
-2 5 3 -6

slope m = (y2-y1)/(x2-x1)
( -6 - 5 )/( 3 - -2 )
( -11 / 5 )
m= -2 1/5
slope = -11/5
Tan = -11/5
Theta = tan ^-1(11/5)
= -65.55 deg

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cos x =
square both sides
cos^2 x=3/49
1-cos^2 x=1-3/49
sin^2 x = 46/49


cos 2x = 2 cos^2 x -1
=2*3/49-1
=6/49-1
=-43/49
Cos 2x= -43/49
Sin 2x = 2sin x . cos x
=
=
Tan (2x) = sin (2x)/cos (2x)

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option C

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option (c)

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log(3) 1/27=x
3= (1/27)^x







1=-3x
-1/3 =x

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The equation in point slope form is
y-y1=m(x-x1)
m=-1
(x1,y1)=> (2,4)
y-4 = -1(x-2)
y-4=-x+2
y=-x+2+4
y=-x+6

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two planes having the same airspeed depart at the same time, flying in opposite directions. one plane flies into 35mph wind, and the other flies in the same direction as the wind. After a period of time, on plane has traveled 460 m an the other has traveled 404 m. let x = airspeed of each plane ( airspeed is the same for each plane)
a) write an expression for the speed of the plane traveling into the wind.
x-35 mph
b) Traveling with the wind
x+35 mph
c) write an expression for the time the same plane has been in the air when it has traveled 460 m
460/(x+35)
d) what is the relationship for the times the planes have been in the air
Times are equal
second plane time = 404/(x-35)
c) write an equation modeling this relationship.
460/(x+35)=404/(x-35)
460(x-35)=404(x+35)
460x-16100=404x+14140
56x=30240
x=540 mph

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18 feet ladder

75 deg angle
Height of window
Sin 75 = height / ladder length
height = sin 75 * 18
=>0.965*18
=>17.37 feet
height of window =17.37
if ladder length is 16 feet even when kept vertically it will not reach the window.

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Laura received an e-mail from Billy. After 10 minutes Laura forwarded it to 3 of her friends. After 10 minutes each of those 3 friends forwarded the message to 3 more friends. If the message was forwarded like this every 10 minutes, how many people had received Billy's e-mail after 50 minutes?

the email was sent every 10 minutes
Total emails 5
first email 3, 9,27 81, 243.....
it is a GP




Sn = 3*242/2
=363 emails will be sent in 50 minutes

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f(x) = 3x2 + 6x + 1

On X axis y=0
3x^2+6x+1=0
Find the roots of the equation by quadratic formula

a= 3 b= 6 c= 1

b^2-4ac= 36 - 12
b^2-4ac= 24 = 4.9


x1=( -6 + 4.9 )/ 6
x1= -0.18
x2=( -6 -4.9 ) / 6
x2= -1.82

Intersection at -0.18 and -1.82

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If they are moving in same direction they will not meet.
if opposite then it can be worked out

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2 x + 1 y = 2
Find the slope of this line

1 y = -2 x + 2
Divide by 1
y = -2 x + 2
Compare this equation with y=mx+b, m= slope & b= y intercept
slope m = -2

The slope of a line perpendicular to the above line will be the negative reciprocal 1/2
Because m1*m2 =-1
The slope of the required line will be 1/2

m= 1/2 ,point ( 4 , 7 )
Find b by plugging the values of m & the point in
y=mx+b
7 = 2 + b
b= 5
m= 1/2
The required equation is y = 1/ 2 x + 5


where do they intersect
2.00 x + 1.00 y = 2.00 .............1
Total value
-0.50 x + 1.00 y = 5.00 .............2
Eliminate y
multiply (1)by -1.00
Multiply (2) by 1.00
-2.00 x -1.00 y = -2.00
-0.50 x + 1.00 y = 5.00
Add the two equations
-2.50 x = 3.00
/ -2.50
x = -1.20
plug value of x in (1)
2.00 x + 1.00 y = 2.00
-2.40 + y = 2.00
y = 2.00 + 2.40
y = 4.40
y = 4.40

(-1.20,4.40) (6,7)
Find the distance between the points. that's the length of the pipe line
Distance between two points
x1 y1 x2 y2

-1.2 4.4 6 7
d=
d=
d=
d=
d= 7.66
m.ananth@hotmail.ca
m.ananth@hotmail.ca

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Boat speed speed =x mph
current speed =y mph
against current 14 hours
with current 12 hours

Distance against 336 miles distance with 336 miles
t=d/r against current -
336 / ( x - y )= 14
14 ( x - y ) = 336
14 x - 14 y = 336 ....................1

336 / ( x + y )= 12
12 ( x + y ) = 336
12 x + 12 y = 336 ...............2
Multiply (1) by 6
Multiply (2) by 7
we get
84 x + -84 y = 2016
84 x + 84 y = 2352
168 x = 4368
/ 168
x = 26.00 mph

plug value of x in (1) y
14 x -14 y = 336
364 -14 -364 = 336
-14 y = 336
-14 y = -28 mph
y = 2.00
Boat speed speed 26.00 mph
current speed 2.00 mph

m.ananth@hotmail.ca

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First car 40 mph
Car II 65 mph
First car 09:00
Car II 11:00
Difference in time= 02:00 => 2.00 hours
First car will have covered 80.00 miles Car II starts
catch up distance= 80.00 miles
catch up speed = 65 -40
catch up speed = 25 mph
Catchup time = catchup distance/catch up speed
catch up time= 3.2
catch up time= 3.20 hours
They will meet at 12:20 pm
Car II speed = 65 hours
Time to catch up = 3.2
D=Speed * time
D= 65 * 3.2
D= 208 miles
Car II will catch up when it has traveled 208 miles

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we have two unknown variables, the radius (r) and the height (h).
Volume (V) = pir^2h


30 = pir^2h
h = 30/pir2
Substitute this into the surface area equation for h, we get
A = (2pir( 30 )/pir^2) + 2pir2)

A = (60/r) + 2pir2

Take the derivative for r

A = 60r-1 + 2pir2

A' = -60r-2 + 4pir


when slope is zero

0 = -60r-2 + 4pir


60r-2 = 4pir

Multiply both sides by r^2

60 = 4pir3


4.775 = r3

1.68 = r
h = 30/pir^2
h=30/pi*(1.68)^2
h = 3.38


A = 2pirh + 2pir^2
A = 2*pi*(1.68)(3.38) + 2pi*(1.68)^2
A = 35.68 + 17.73
A = 53.41 square inches
CHECK

V= pi*1.68^2*3.38= 29.96
you can check for maxima or minima by taking the second derivate for more detailing

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4.10,Rs.3.75 and 4.50 per kg
suppose he mixes 5 kg,4kg & 1 kg of mix to make 10 kg
5*4.10+4*3.75+1*4.5=40
Mixture costs 40/10 = Rs.4.00 /kg
4*1.25 = Rs. 5 /kg

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(3-x)+(6)+(7-5x) is a geometric series, find two possible values for
a) x
b) the common ratio
c) the sum of the GP.

(3-x)(7-5x)=6^2
21-15x-7x+5x^2=36
5x^2-22x-15=0
5x^2-25x+3x-15=0
5x(x-5)+3(x-5)=0
(x-5)(5x+3)=0
x=5 OR -3/5
if x=5
-2,6,-28
common ratio = -4
you should be able to continue

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put it in the right format

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=

=


=
=0




since the index is odd
x=1

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canoe speed x mph
current speed 2 kph

against current x- 2 mph
with current x+ 2 mph

Distance= 8 miles

Time against + time with = 5 1/3 hours
t=d/r

8 /( x + 2 ) + 8 /(x - 2 ) = 5 1/3

LCD = (x - 2 ) ( x + 2 )
8 *( x - 2 ) + 8 (x + 2 ) = 5 1/3 (x^2 - 4 )
8 x - -16 + 8 x + 16 = 5 1/3 ( x ^2 - 4 )
16 x = 5 1/3 x ^2 -21.33
5 1/3 x ^2 - -16 x - 21.33

Find the roots of the equation by quadratic formula

a= 5 1/3 , b= -16 , c= -21.33

b^2-4ac= 256 + 455.04
b^2-4ac= 711.04

0
1.33
x1=( 16 + 26.67 )/ 10.67
x1= 4.00

x2=( 16 -26.67 ) / 10.67
x2= -1.00
Ignore negative value
canoe speed 4.000 mph

m.ananth@hotmail.ca

canoe speed x mph
current speed 2 kph

against current x- 2 mph
with current x+ 2 mph

Distance= 8 miles

Time against + time with = 5 1/3 hours
t=d/r

8 /( x + 2 ) + 8 /(x - 2 ) = 5 1/3

LCD = (x - 2 ) ( x + 2 )
8 *( x - 2 ) + 8 (x + 2 ) = 5 1/3 (x^2 - 4 )
8 x - -16 + 8 x + 16 = 5 1/3 ( x ^2 - 4 )
16 x = 5 1/3 x ^2 -21.33
5 1/3 x ^2 - -16 x - 21.33

Find the roots of the equation by quadratic formula

a= 5 1/3 , b= -16 , c= -21.33

b^2-4ac= 256 + 455.04
b^2-4ac= 711.04

0
1.33
x1=( 16 + 26.67 )/ 10.67
x1= 4.00

x2=( 16 -26.67 ) / 10.67
x2= -1.00
Ignore negative value
canoe speed 4.000 mph

m.ananth@hotmail.ca


canoe speed x mph
current speed 2 kph

against current x- 2 mph
with current x+ 2 mph

Distance= 8 miles

Time against + time with = 5 1/3 hours
t=d/r

8 /( x + 2 ) + 8 /(x - 2 ) = 5 1/3

LCD = (x - 2 ) ( x + 2 )
8 *( x - 2 ) + 8 (x + 2 ) = 5 1/3 (x^2 - 4 )
8 x - -16 + 8 x + 16 = 5 1/3 ( x ^2 - 4 )
16 x = 5 1/3 x ^2 -21.33
5 1/3 x ^2 - -16 x - 21.33

Find the roots of the equation by quadratic formula

a= 5 1/3 , b= -16 , c= -21.33

b^2-4ac= 256 + 455.04
b^2-4ac= 711.04

0
1.33
x1=( 16 + 26.67 )/ 10.67
x1= 4.00

x2=( 16 -26.67 ) / 10.67
x2= -1.00
Ignore negative value
canoe speed 4 mph

m.ananth@hotmail.ca

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option (c)
the slopes are negative reciprocals
slope of given line is 1/3
slope of perpendicular line will be -3

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here you have to use the law of sines
let tree make an angle of 83 deg.with the ground. (YZ)angle ZYX = 83deg
he walks 100 feet away from tree to point X (YX)angle YXZ= 33 deg
so remaining angle YZX = 64 deg
triangle YXZ is formed
sin A /a = sin B /b = sin C /c
Sin YZX/100 = sin YXZ/ YZ
sin 64/100 = sin 33/YZ
100 * sin 33/sin 64 = YZ
100*0.54/0.898= YZ
YZ= 60.13 feet the height of the tree
You can solve the problem in a similar way if the tilt is to the other side.

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Boat speed =x kph
current speed =y kph
against current 1.5 hours
with current 1 hours

Distance against 24 miles distance with 24 miles
t=d/r against current -
24.00 / ( x - y )= 1.50
1.50 ( x - y ) = 24.00
1.50 x - 1.50 y = 24.00 ....................1

24.00 / ( x + y )= 1.00
1.00 ( x + y ) = 24.00
1.00 x + 1.00 y = 24.00 ...............2
Multiply (1) by 1.00
Multiply (2) by 1.50
we get
1.50 x + -1.50 y = 24.00
1.50 x + 1.50 y = 36.00
3.00 x = 60.00
/ 3.00
x = 20.00 mph

plug value of x in (1) y
1.50 x -1.50 y = 24.00
30.00 -1.50 -30.00 = 24.00
-1.50 y = 24.00
-1.50 y = -6.00 kph
y = 4.00
Boat speed 20.00 mph
current speed 4.00 kph

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let speed of current be c kph
upstream speed = 10-c
upstream distance = 12
down stream speed = 10+c
down stream distance = 28 km
upstream time + downstream time = 4
t=d/r
12/(10-c) + 28/(10+c)= 4
multiply the equation by (10+c)(10-c)

12(10+c)+28(10-c)=4(10+c)(10-c)
120+12c+280-28c=4(100-c^2)
400-16c=400-4c^2
re arrange
4c^2=16c
/4c
c=4

current speed = 4 kph