Answer:
The statements are incorrect as: The sum of even numbers from 1 to 100(i.e. 2550) is not double\twice of the sum of odd numbers from 1 to 100(i.e. 2500).
Step-by-step explanation:
We know that sum of an Arithmetic Progression(A.P.) is given by:
where 'n' denotes the "number" of digits whose sum is to be determined, 'a' denotes the first digit of the series and '' denote last digit of the series.
Now the sum of even numbers i.e. 2+4+6+8+....+100 is given by the use of sum of the arithmetic progression since the series is an A.P. with a common difference of 2.
image with explanation
Hence, sum of even numbers from 1 to 100 is 2550.
Also the series of odd numbers is an A.P. with a common difference of 2.
sum of odd numbers from 1 to 100 is given by: 1+3+5+....+99
.
Hence, the sum of all the odd numbers from 1 to 100 is 2500.
Clearly the sum of even numbers from 1 to 100(i.e. 2550) is not double of the sum of odd numbers from 1 to 100(i.e. 2500).
Hence the statement is incorrect.
Step-by-step explanation:
Answer:
see explanation
Step-by-step explanation:
Using the trigonometric identity
sin²x + cos²x = 1 , then
sin²x = 1 - cos²x and cos²x = 1 - sin²x
Consider the left side
← expand numerator/denominator using FOIL
=
= 
= 
= 
= cot²A = right side , thus proven
Let t, h, b represent the weighs of tail, head, and body, respectively.
t = 4 . . . . given
h = t + b/2 . . . . the head weighs as much as the tail and half the body
b/2 = h + t . . . . half the body weighs as much as the head and tail
_____
Substituting for b/2 in the second equation using the expression in the third equation, we have
... h = t + (h + t)
Subtracting h from both sides gives
... 0 = 2t . . . . . . in contradiction to the initial statement about tail weight.
Conclusion: there's no solution to the problem given here.
Undefined!! i hope this helps
It's 192, I'm pretty sure. I hope that you get this right! :^)