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?
@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 ?
the condition for the roots of the equation PX^3+3QX^2+3RX+S=0 to be in geometric mean is?
1. EF^2=1 or E^2F=1
2. E=F^2 or F=E^2
3. EF^3=1 or E^3F=1
4. E=F^3 or F=E^3
f(x) = x^3 - 6x^2 + 14x - 4
Thus, f(4) = 20
@vbhvgupta said:the condition for the roots of the equation PX^3+3QX^2+3RX+S=0 to be in geometric mean is?
SQ^3 = PR^3
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?
@vbhvgupta said:If E and F are the root of the equation ax^2+bx+c=0 where c^3+abc+a^3=0 which of the following is true1. EF^2=1 or E^2F=12. E=F^2 or F=E^23. EF^3=1 or E^3F=14. E=F^3 or F=E^3
substitute a=1,c=2 => b =-4.5
2x2 - 9x + 4
(2x - 1)(x - 4) => E= 0.5 , F = 4 or E=4 , F=0.5
Answer - option 1 ?
@Tusharrr said:@techgeek2050 How
u just need to draw the figure clearly
let the point of contact of the semi circle and triangle be X and Y
center of circle be O
r be the radius
AX will be 24 - r
OY will be r
triangles AXO and OYC are similar
so
r/(7 - r) = (24 - r)/r
r = 168/31
a + b = 168 + 31 = 199
@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?
E+F+1 = 19?
@vbhvgupta said:If E and F are the root of the equation ax^2+bx+c=0 where c^3+abc+a^3=0 which of the following is true1. EF^2=1 or E^2F=12. E=F^2 or F=E^23. EF^3=1 or E^3F=14. E=F^3 or F=E^3
b=0
a=1 and c=-1
x^2 - 1 = 0
x^2 = 1
=> e/f = +/-1
a=1 and c=-1
x^2 - 1 = 0
x^2 = 1
=> e/f = +/-1
ye toh koi option definitely true nahi ho rahi....koi aur combination try karna padega 
@vbhvgupta said:the condition for the roots of the equation PX^3+3QX^2+3RX+S=0 to be in geometric mean is?
Let roots be a/r, a and ar
Then, -S/P = a^3
And, a(1/r + 1 + r) = -3Q/P
a^2(1 + r + 1/r) = 3R/P
Dividing above two, a = -R/Q
a^3 = -(R/Q)^3 = -S/P
Hence, R^3/Q^3 = S/P ?
The sun goes around the center of our galaxy once every 250 million years. The sun is also 2.55 —1020 m from the center of our galaxy. What is the acceleration of our sun towards the center of the galaxy in m/s2?
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.
The 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 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:3/2
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)/2
R= 2r +/-2 rt r^2-4
twice of 8 or 2 must be equal to this root.
taking 2 .
4-2r = 2rt r^2-4
2-r = rt R^2-4
sq it
8 = 4r
r=2.
correct me if i am wrong. :)
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)/2
R= 2r +/-2 rt r^2-4
twice of 8 or 2 must be equal to this root.
taking 2 .
4-2r = 2rt r^2-4
2-r = rt R^2-4
sq it
8 = 4r
r=2.
correct me if i am wrong. :)