12 distinct edges => 12 distinct mid points
12 face diagonals = > 6 distinct mid points, one for each face
4 body diagonals => 1 distinct mid point (centre of cube)
28 line segments => 19 distinct mid points
@Tusharrr 3 types of line segments are obtained...edges, face diagonals, body diagonals
@Tusharrr said:A cube has 8 vertices. For each pair of distinct vertices, we connect them up with a line segment. There are 28 such line segments. For each of these 28 line segments, we mark the midpoint. How many distinct points have been marked as the midpoints?
1 on each face ==> 6
1 on each edge ==> 12
1 at the center ==> 1
total = 19
@Tusharrr said:A cube has 8 vertices. For each pair of distinct vertices, we connect them up with a line segment. There are 28 such line segments. For each of these 28 line segments, we mark the midpoint. How many distinct points have been marked as the midpoints?
19 ?
6 points at the center of each face
12 at the edges
1 at the center of cube
6 + 12 + 1 = 19
@Tusharrr said:ABC is a right angled triangle with ˆ ABC=90 ˆ˜and side lengths AB=24 and BC=7. A semicircle is inscribed in ABC, such that the diameter is on ACand it is tangent to AB and BC. If the radius of the semicircle is an improper fraction of the form a/b, where a and b are positive, coprime integers, what is the value of a+b?
199.
@Tusharrr said:ABC is a right angled triangle with ˆ ABC=90 ˆ˜and side lengths AB=24 and BC=7. A semicircle is inscribed in ABC, such that the diameter is on ACand it is tangent to AB and BC. If the radius of the semicircle is an improper fraction of the form a/b, where a and b are positive, coprime integers, what is the value of a+b?
r/24 = oc/25 ( o is the centre)
r/7= ao/25 ( similar triangles)
r = 168/31 = a/ b
199
Plz solve
Details and assumptions
You may assume the sun's orbit is circular.There are 24 hours in a day and 365 days in a year.
@vbhvgupta said:If one of the root of the equation x^2-10x+16=0 is half of one of the root of the equation x^2-4Rx+16=0 Find R such that both the equation have integral roots
x^2-10x+16=0 , root1 = 8, root2 =2
For x^2-4Rx+16=0 ,
root1 + root2 = 4R
root1 * root2 = 16
=> (root1, root2) = (1,16), (2,8), (4,4)
Possible solutions should have 4(double of 2) or 16(double of 8) as one of the root.
If one root of eq1 is equal to half of eq2's exactly one root then, solution is (1,16) and 4R=17, R = 17/4.
Else R = 2 or 17/4 ??
@vbhvgupta said:
the condition for the roots of the equation PX^3+3QX^2+3RX+S=0 to be in geometric mean is?
P=(-Q)=R=S 
@ChirpiBird said:sry, didnt get you? answer is 3/2?this is how i did,roots of the first eq are 8,2.for next eq. roots will be R = (+ 4r +/- rt 16r^2- 64)/2R= 2r +/-2 rt r^2-4twice of 8 or 2 must be equal to this root.taking 2 . 4-2r = 2rt r^2-42-r = rt R^2-4sq it8 = 4rr=2.correct me if i am wrong.
will post the solution in evening..in office now.
two equation have a common root which is positive. The other root of the equation satisfy x^2-9x+18=0 The product of the sum of the root of the two equation is 40. Find the common root.
The root of the equation X^3+PX^2+QX+R=0 are consecutive positive integer. Which of the following can never be the value of Q
1. 47
2. 11
3. 27
4. 107
1. 47
2. 11
3. 27
4. 107
If the equation X^3-4X^2+X+6=0 and X^3-3X^2-4X+K=0 have a common root. which of the following be the value of k
1. 12
2. 6
3. -12
4. -3
1. 12
2. 6
3. -12
4. -3
@Tusharrr said:Plz solveThe sun goes around the center of our galaxy once every 250 million years. The sun is also 2.55 —10^20 m from the center of our galaxy. What is the acceleration of our sun towards the center of the galaxy in m/s^2?Details and assumptionsYou may assume the sun's orbit is circular.There are 24 hours in a day and 365 days in a year.
R = 2.55 —10^20 m
Time period T = 250 x 10^6 x 365 x 24 x 60 x 60 seconds
angular velocity w = 2pi /T
centripetal acceleration = w^2 x R
put the values now
@anshiiii said:x^2-10x+16=0 , root1 = 8, root2 =2For x^2-4Rx+16=0 , root1 + root2 = 4Rroot1 * root2 = 16=> (root1, root2) = (1,16), (2,8), (4,4)Possible solutions should have 4(double of 2) or 16(double of 8) as one of the root.If one root of eq1 is equal to half of eq2's exactly one root then, solution is (1,16) and 4R=17, R = 17/4.Else R = 2 or 17/4 ??
ans is 3/2 i think
@vbhvgupta said:two equation have a common root which is positive. The other root of the equation satisfy x^2-9x+18=0 The product of the sum of the root of the two equation is 40. Find the common root.
2 ? not sure...
@Tusharrr said:Plz solveThe sun goes around the center of our galaxy once every 250 million years. The sun is also 2.55 —10^20 m from the center of our galaxy. What is the acceleration of our sun towards the center of the galaxy in m/s^2?Details and assumptionsYou may assume the sun's orbit is circular.There are 24 hours in a day and 365 days in a year.
Acceleration = v^2/r
= (2(PI)r/t)^2 / r t is time taken to complete one revolution
@vbhvgupta said:If the equation X^3-4X^2+X+6=0 and X^3-3X^2-4X+K=0 have a common root. which of the following be the value of k
1. 12
2. 6
3. -12
4. -3
12
@vbhvgupta said:two equation have a common root which is positive. The other root of the equation satisfy x^2-9x+18=0 The product of the sum of the root of the two equation is 40. Find the common root.
Let common root be a
x^2 - 9x + 18 = 0, roots = 3, 6
Product of sum of roots = (3+a)(6+a) = 40 = 5*8
=> a = 2