A fast moving superfast express crosses another pasenger train in 20 seconds. The speed of faster train is 72 km/hr and speeds of slower train is 27 km/h. Also the length of faster ntrain is 100m, then find the length of the slower train if they are moving in the same direction.
Time=Sum of length of the two trainsDifference in speeds
20=(100+x)252
⇒ X=150m
It is given that log10 2= 0.301 and log10 3 = 0.477. How many digits are there in (108)10?
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If 5x - 17 = -x + 7, then x = ?
Given equation 5x-17 = -x+7
Add 1x to each side of the equation
5x-17+x = -x+7+x
6x=24
x=4
Solve the equation 122x+1 = 1 ?
Rewrite equation as 122x+1 = 120
Leads to 2x + 1 = 0
Solve for x : x = -1/2
If log72 = m, then log4928 is equal to ?
log4928 = 12log77×4
= 12+122log72= 12+log72= 12 + m= 1+2m2.
If a2+b2 = c2 , then 1logc+ab + 1logc-ab = ?
Given a2 + b2 = c2
Now 1logc+ab + 1logc-ab
= logbc+a + logbc-a
= logbc2-a2
= 2logbb = 2
If log 64 = 1.8061, then the value of log 16 will be (approx)?
Given that, log 64 = 1.8061
i.e log43=1.8061
--> 3 log 4 = 1.8061
--> log 4 = 0.6020
--> 2 log 4 = 1.2040
⇒log42=1.2040
Therefore, log 16 = 1.2040
For x∈N, x>1, and p=logxx+1, q=logx+1x+2 then which one of the following is correct?
kl>k+1l+1 for (k,l) > 0 and k > l
Let k = x+1 and l = x
Therefore, x+1x>(x+1)+1(x)+1
(x + 1) > x
Therefore, log(x+1)log(x)>log(x+2)log(x+1)
⇒logxx+1 >logx+1x+2
The Value of logtan10+logtan20+⋯⋯+logtan890 is
= log tan10+log tan890 + log tan20+ log tan880+⋯⋯+log tan450
= log [tan10 × tan890] + log [tan20 × tan880 ] +⋯⋯+log1
∵ tan(90-θ)=cotθ and tan 450=1
= log 1 + log 1 +.....+log 1
= 0.