ABC is an acute triangle with ∠BCA=35∘. Denote the circumcenter of ABC as O and the orthocenter of ABC as H. If AO=AH, what is the value of ∠ABC(in degrees)?
Details and assumptions
The circumcenter of a triangle is the center of a circle which passes through all three vertices of a triangle. The orthocenter of a triangle is the intersection of the 3 altitudes (perpendicular from vertices to opposite side).
Answer is integer
@ananyboss said:Hi,Need to solve this using allegation In a zoo their are deers and ducks. 180 heads and 448 legs. How many deers are their ?
44
@amresh_maverick said:Let T be the set of integers {2, 12, 22, 32 ..., 542, 552} and S be a subset of T such that the sum of no two elements of S is divisible by 3. The maximum possible number of elements in S is
OA: 20
@amresh_maverick said:N= 1^1 × 2^2 × 3^6 × 4^12 × 5^20 ..... 25 terms. Find the highest power of 75 that can divide N.
OA 950
@vbhvgupta said:The duration of a railway journey varies as the distance and inversely as the velocity; the velocity varies directly as the square root of the quantity of coal used per km, and inversely as the no of carriages in the train. In a journey of 50 KM in half an hour with 18 carriages, 100 kg of coal is required. how much coal will be consumed in a journey of 42 KM in 28 minutes with 16 carriages.
T~d....T~1/v..T=kd/v
v~[c]^1/2....v~1/carriage....v=k(1)[c]^1/2/carriage
T=kd/k(1)[c]1/2/carriage=k/k(1){d*carriage/[c]^1/2}
1/2hr.=k/k(1)*50*18/100^1/2
=>k/k(1)=1/180
28/60=1/180*42*16/[c]^1/2=64..
section of the 3 altitudes (perpendicular from vertices to opposite side).
How many digits cannot be the unit €™s digit of the product of 3 three-digit numbers whose sum is 989?
Options- 2,4,1,3
@shattereddream said:How many digits cannot be the unit €™s digit of the product of 3 three-digit numbers whose sum is 989? Options- 2,4,1,3
3?
2 3 and 7?
@shattereddream said:How many digits cannot be the unit €™s digit of the product of 3 three-digit numbers whose sum is 989? Options- 2,4,1,3
Is it 3? Not sure tho
@shattereddream said:How many digits cannot be the unit €™s digit of the product of 3 three-digit numbers whose sum is 989? Options- 2,4,1,3
3??
@shattereddream said:How many digits cannot be the unit €™s digit of the product of 3 three-digit numbers whose sum is 989? Options- 2,4,1,3
4??
No.'s are 1,3,8,9
No.'s are 1,3,8,9
@shattereddream said:How many digits cannot be the unit €™s digit of the product of 3 three-digit numbers whose sum is 989? Options- 2,4,1,3
sum of the units' digit should be 9
now for 2 . 1 + 6 + 2 = 9 // now the unit of product would end in 2
for 1, we have 9 + 9 + 1 = 19... so product unit's digit = 1
now 4 can be written as = 6*6*7 sum of which is 19
so 3 is the answer....
[image was removed because it was too large]
@Logrhythm said:hahahah....i remember i did that in an interview when i asked "what is your fav subject" i replied "math" and uske baad it was a colossal fiasco.....pata nahi usko laga meine math bola toh mein Aryabhatt hun ok, no spammingq) what is the highest power of 12 that divides (5^36 - 1) 1,2,3,4
5^36-1= 25^18-1
=(24+1)^18-1=18c0 24^18*1^0+ 18c1 24^17 *1 ^1+.................+18c17 24 + 18c18*1-1
now thsi whole expression is divisible by 12 but the question is what will be the highest power
obviously the highest will be the one that divides 18c17 *24
18c17*24=18*24/.the highest power dividing this is 2
=(24+1)^18-1=18c0 24^18*1^0+ 18c1 24^17 *1 ^1+.................+18c17 24 + 18c18*1-1
now thsi whole expression is divisible by 12 but the question is what will be the highest power
obviously the highest will be the one that divides 18c17 *24
18c17*24=18*24/.the highest power dividing this is 2
There are two trains - A and B, both starting from point S, with train A starting 10 seconds after train B. The length and speed for train A and train B are 100 m, 270 kmph and 200 m, 216 kmph respectively. How much time after start will train A reach train B €™s tail?
@amresh_maverick said:There are two trains - A and B, both starting from point S, with train A starting 10 seconds after train B. The length and speed for train A and train B are 100 m, 270 kmph and 200 m, 216 kmph respectively. How much time after start will train A reach train B €™s tail?
speed of train A=270 *1000/3600 = 75 m/s
speed of train B=60
in ten seconds train B would have travelled 600 m
tail would have travelled 600-200=400m
400/15 = 80/3 seconds = 26.67 seconds