@chillfactor said:1+sin(pi/4+k)+2cos(pi/4-k)= 1 + (1/√2)(Sink + Cosk) + √2(Cosk + Sink)Max = 1 + (√2)(1/√2) + (√2)(√2) = 4
sir how to write this..bold part
Again it has hpnd..an empty post..is it because im using windows phone or smthng...anyws hers a qstn...x5-10x4 plus ax3 plus bx2 plus cx-32 equals 0 . All roots of above eqn R positive, find a plus b plus c.
Options---
0, 80,160,240, none....no OA.
@manasvr said:Again it has hpnd..an empty post..is it because im using windows phone or smthng...anyws hers a qstn...x5-10x4 plus ax3 plus bx2 plus cx-32 equals 0 . All roots of above eqn R positive, find a plus b plus c.Options---0, 80,160,240, none....no OA.
EDIT...roots are real and positive...how do I delete or edit my posts..
@19rsb said:The max value of 1+sin(pi/4+k)+2cos(pi/4-k) for real values of k isa)3 b)5c)4 d)none of these
sin (pi/4+k)=1/_/2*cos k+1/_/2*sin k=1/_/2*(sin k+cos k)
cos (pi/4-k)=1/_/2*cos k+1/_/2*sin k=1/_/2*(sin k+cos k)
So, 1+sin(pi/4+k)+2cos(pi/4-k) =1+ 1/_/2*(sin k+cos k)+ _/2*(sin k+cos k)
maximum value of sin k+cos k=_/2
=> maximum value of expression=1+1/_/2*_/2+_/2*_/2=1+1+2=4
@manasvr said:Again it has hpnd..an empty post..is it because im using windows phone or smthng...anyws hers a qstn...x5-10x4 plus ax3 plus bx2 plus cx-32 equals 0 . All roots of above eqn R positive, find a plus b plus c.Options---0, 80,160,240, none....no OA.
At rightmost corner, there is an option of deleting and editing of any post :)
sum of the roots=10
multiplication of roots=32 and all roots are real and positive
==> roots may be 2,2,2,2,2
==> a=40, b=-80, c=80 => a+b+c=40?
question thoda shi se post karo :/
@19rsb said:The max value of 1+sin(pi/4+k)+2cos(pi/4-k) for real values of k isa)3 b)5c)4 d)none of these
all we have to do in this question is to apply the simple replation: cos(90-x) = sinx
1+ sin(45+k) + 2cos(45-k)
=1 + sin(45+k) + 2cos{90 - (45+k)}
=1 + sin(45+k) + 2sin(45+k)
=1 + 3sin(45+k)
now we know that maximum value of a sin function is 1. Hence max possible value of
1+ 3sin(45+k) = 1 + 3
= 4.
ATDH.
@manasvr said:Again it has hpnd..an empty post..is it because im using windows phone or smthng...anyws hers a qstn...x5-10x4 plus ax3 plus bx2 plus cx-32 equals 0 . All roots of above eqn R positive, find a plus b plus c.Options---0, 80,160,240, none....no OA.
x^5-10x^4 + ax^3 + bx^2 + cx-32 = 0
pqrst
sum of roots = 10
pqrst = 32 = 2.2.2.2.2
then eqn is
(x-2)^5 = 0
x^5-10x^4 + 40x^3 - 80x^2 +80x-32 = 0
a+b+c = 40
a student was given 8 two-digit numbers to add,by a teacher.If the student reversed each number and added the results,the sum of the numbers would be 36 more than the actual sum.Find the excess of the sum of the units digits over that of the ten digits.
@meenu05 said:In a triangle Abc length of Ab and Ac is 8 . The radius of the incircle drawn in the triangle Abc with the enter o is 2 and the length of Ao is 4 cm . Find the length of the side Bc ?
@anantn said:4 root 7
@tani90 said:no, it would be 4_/7, I forgot to double it before....
ye kaise aaya??
mera toh 4*(4-rt(3)) aa rha hai...
:(
@nole said:a student was given 8 two-digit numbers to add,by a teacher.If the student reversed each number and added the results,the sum of the numbers would be 36 more than the actual sum.Find the excess of the sum of the units digits over that of the ten digits.
4
:)
@pari19O1 yeah correct answer.They have asked the excess of the sum of the unit digits over that of the ten digits.though i got the logic,but how do we know that if they are refering to the original number or reversed number in the last line?
@nole said:@pari19O1 yeah correct answer.They have asked the excess of the sum of the unit digits over that of the ten digits.though i got the logic,but how do we know that if they are refering to the original number or reversed number in the last line?
it shud be for the original numbers....
cause that is the number that was given by the teacher....
student added the reverse but those were not the numbers...
:)
@nole said:a student was given 8 two-digit numbers to add,by a teacher.If the student reversed each number and added the results,the sum of the numbers would be 36 more than the actual sum.Find the excess of the sum of the units digits over that of the ten digits.
10a1+b1 + 10a2 + b2 + ... + 10a8 + b8 = X
10b1 + a1 + ........................+ 10b8 + a8 = 36 + X
subtracting.
9b1 - 9 a1 + ....... 9b8 - 9 a8 = 36
9(b1 + b2 + b3 + ...b8 - (a1 + ...+a8)) = 36
so,
(b1 + b2 + b3 + ...b8 - (a1 + ...+a8))=4 ??
A student was given 8 two-digit numbers to add,by a teacher.If the student reversed each number and added the results,the sum of the numbers would be 36 more than the actual sum.Find the excess of the sum of the units digits over that of the ten digits.
cc: @nole
Let us say that the 8 numbers are a1 b1, a2 b2, a3 b3 ... a8 b8 ...{eg: 38}
They actually represent a1*10 + b1, a2*10 + b2 ... a8*10 + b8 ...{eg: 3*10 + 8}
Sum of the numbers = (a1 + a2 + a3 .. a8)*10 + (b1 + b2 + b3 .. b8)
Reversed numbers will be b1 a1, b2 a2, ... b8 a8
Sum of the reversed numbers will be (b1 + b2 + b3 .. b8)*10 + (a1 + a2 + a3 .. a8)
Reverse sum - actual sum = 36 ... {Given in the question}
(b1 + b2 + b3 .. b8)*9 - (a1 + a2 + a3... a8)*9 = 36
=> (b1 + b2 + b3 ... b8) - (a1 + a2 + a3 ... a8) = 4 {9 can be taken common and cancelled from both sides}
=> Sum of the units digits - Sum of the tens digits = 4
@nole said:a student was given 8 two-digit numbers to add,by a teacher.If the student reversed each number and added the results,the sum of the numbers would be 36 more than the actual sum.Find the excess of the sum of the units digits over that of the ten digits.
4
A pole situated on the ground is supported by two wires. One end of each wire is attached to the top of the
pole and other end is fixed to the ground at the points P1 and P2. The distance between points P1and P2 is 20
meters and the line joining P1 and P2 does not pass through the foot of the pole. The wires fixed to P1 and
P2 make angles 30°and 60° with the ground respectively. The height of the pole is 'h' meters. Which of the
following statements is true?
A. 5 3 ≤ h ≤10 3
B. 10 C. 5 3 D. 10 ≤ h ≤ 20
E. None of these
pole and other end is fixed to the ground at the points P1 and P2. The distance between points P1and P2 is 20
meters and the line joining P1 and P2 does not pass through the foot of the pole. The wires fixed to P1 and
P2 make angles 30°and 60° with the ground respectively. The height of the pole is 'h' meters. Which of the
following statements is true?
A. 5 3 ≤ h ≤10 3
B. 10 C. 5 3 D. 10 ≤ h ≤ 20
E. None of these
There is a park which is in the shape of a right-angled triangle. A tower is situated at the mid-point of the largest
edge of the park. The two perpendicular edges of the park subtend angles of 60°and 90° at the top of the tower.
Area of the park is 2592*root(2) m^2.
What is the height of the tower?
A. 72 2 m
B. 36 2 m
C. 36 m
D. 72 m
E. None of these
edge of the park. The two perpendicular edges of the park subtend angles of 60°and 90° at the top of the tower.
Area of the park is 2592*root(2) m^2.
What is the height of the tower?
A. 72 2 m
B. 36 2 m
C. 36 m
D. 72 m
E. None of these
There is an insect at the vertex where perpendicular edges of the park meet and wants to crawl its way to the
foot of the tower. Find the minimum time (in minutes) for which the insect has to trek in order to reach the
foot of the tower, if it can only crawl along the edges of the park with speed of 20×10^(−2) m/s.
foot of the tower. Find the minimum time (in minutes) for which the insect has to trek in order to reach the
foot of the tower, if it can only crawl along the edges of the park with speed of 20×10^(−2) m/s.
A. 3(2 +rt 3 )
B. 3(1+ rt 2 )
C. 12
D. 3(3+ 2 rt2 )
E. 18
B. 3(1+ rt 2 )
C. 12
D. 3(3+ 2 rt2 )
E. 18
@sujamait said:Sir 2nd angle 90 hi hai?There is a park which is in the shape of a right-angled triangle. A tower is situated at the mid-point of the largestedge of the park. The two perpendicular edges of the park subtend angles of 60째and 90째 at the top of the tower.
@sujamait said:A pole situated on the ground is supported by two wires. One end of each wire is attached to the top of thepole and other end is fixed to the ground at the points P1 and P2.
B. 10