for the given A.P
5,9,13,...
First term (a)=5
common difference (d) = 9-5=4
Let Tn=2012
a +(n-1) d=2012
5 +(n-1) 4=2012
(n-1) 4=2012-5
(n-1)4=2007
n-1=501.75
n=1+501.75
n=502.75 -- - - - - -> which is not possible.
No.of terms can never in fraction.
Hence, 2012 is not a term of given A.P
Answer:
24 ft^2
Step-by-step explanation:
just multiply 4 by 3 by 2
Brainilist maybe?
Answer:
5
Step-by-step explanation:
=>
; [∵ 25÷5=5]
=>1*5
=>5
Answer with Step-by-step explanation:
We are given that the set of vectors
is lineraly dependent set .
We have to prove that the set 
is linearly dependent .
Linearly dependent vectors : If the vectors 
are linearly dependent therefore the linear combination
Then ,there exit a scalar which is not equal to zero .
Let 
then the vector 
will be zero and remaining other vectors are not zero.
Proof:
When 
are linearly dependent vectors therefore, linear combination of vectors of given set
By definition of linearly dependent vector
There exist a scalar which is not equal to zero.
Suppose 
then 
The linear combination of the set
When 
Therefore,the set is linearly dependent because it contain a vector which is zero.
Hence, proved .
Answer:
diameter = 2 * radius = 2 * 14 = 28 cm
