Let p (x) = 4x^4-13x^2-2x
The zero of x-2 is 2
putting x = 2 in p (x), we get,
p (2) = 4×2^4-13×2^2-2×2
= 64 - 52 - 4
= 64 - 56
= 8
Therefore, remainder = 8
Let the original number be x
X^2 + 11 = 36
Subtract 11
X^2 = 25
Square root both
X = 5
Solution: her original number is 5
Use BODMAS or BIDMAS (Brackets, over Indices, Division, Multiplication, Addition, the Subtraction, this tells you which order to do things in)
As you cant do √10 in the brackets you do the indices, so (√10)³
Split this up to make it easier
(√10)³= √10 x √10 x √10 = 10√10
You the multiply this by 9
9 x 10√10 = 90√10
then multiply by 5
5 x 90√10 = 450√10 = 1423.024947 (using calculator)
Answer:
Is this a question. If so I will change my answer
Step-by-step explanation:
