II. f(x) doubles for each increase of 1 in the x values. Thus, r must be 2, and so we our ar^1 = 6 from ( I ) above becomes f(x) = a*2^x. Applying the restriction ar^1 = 6 results in f(1) = a*2^1 = 6, or a = 3.
Then f(x) = ar^x becomes f(x) = 3*2^2 (Answer A)
Answer: i) 30 students for chemistry
ii) 21 students for only one subject
iii) 61 students
Step-by-step explanation: from the question we got 30 students registered for chemistry. number students only for biology = (27-18) = 9
number of students only for chemistry = (30-18) = 12
number of student registered for only one subject = (12+9)= 21
number of students registered for neither two subjects = {100-(9+12+18)} = 61
Each number in the sum is even, so we can remove a factor of 2.
2 + 4 + 6 + 8 + ... + 78 + 80 = 2 (1 + 2 + 3 + 4 + ... + 39 + 40)
Use whatever technique you used in (a) and (b) to compute the sum
1 + 2 + 3 + 4 + ... + 39 + 40
With Gauss's method, for instance, we have
S = 1 + 2 + 3 + ... + 38 + 39 + 40
S = 40 + 39 + 38 + ... + 3 + 2 + 1
2S = (1 + 40) + (2 + 39) + ... + (39 + 2) + (40 + 1) = 40×41
S = 20×21 = 420
Then the sum you want is 2×420 = 840.
The answer is 15w^2 + 225w
Key:
^ - Means that it is an exponent, in case you didn't know :)
Let the two sides of a right triangle be equal to one, which means that the hypotenuse is √2
Since cosa=adjacent side / hypotenuse
cos45=1/√2
We can rationalize the denominator by multiplying numerator and denominator by √2
√(2)/2
or if you prefer: √(1/2)
