(2n+1)(2n-1)(n+5)
=(4n^2+2n-2n-1)(n+5)
=(4n^2-1)(n+5)
=4n^3-n+20n^2-5
4n^3+20n^2-n-5
Given, the ratio of blocks A, B, C,D are in the ratio 4:7:3:1
Let us consider the common ratio to be ‘x’.
So, toy blocks with alphabet A is 4x and
toy blocks with alphabet B is 7x and
toy blocks with alphabet C is 3x and
toy blocks with alphabet D is x
Again, the number of ‘A’ blocks is 50 more than the number of ‘C’ blocks.
As no. of ‘A’ and ‘C’ blocks are 4x and 3x respectively.
So,
4x=50 + 3x
x=50
Thus, the number of ‘B’ blocks is 7x = 7(50) = 350
350 is the required number.
Answer:
8 and 5
Step-by-step explanation:
You could rewrite
as
and be tempted to cancel out the factors of
. But this cancellation is only valid when
.
When
, you end up with the indeterminate form
, which is why
is not a zero.
