The 400th term is 425.There are floor(√400) = 20 squares in the range 1..400, so the 400th term will be at least 420. There are floor(∛420) = 7 cubes in the range 1..400, so the 400th term may be as high as 427. However, there are
![\lfloor\sqrt[6]{427}\rfloor=2](https://tex.z-dn.net/?f=%5Clfloor%5Csqrt%5B6%5D%7B427%7D%5Crfloor%3D2)
numbers that are both squares and cubes. Consequently, the 400th term will be 427-2 =
425.
The easy way to think about this is what number comes between 5 and 7? The answer is 6 (but we have to consider that it's negative in this case.) So the answer is -6 Another way to look at it is with a number line (not to "Which number is greater than -7 and less than -5"? scale.)<===(-7)==(-6)==(-5)==(-3)==(-2)==(-1)==(0)==(1)==>Notice how -6 comes between the two numbers. -6 is larger than -7 (it's "less negative" -- "Which number is greater than -7 and less than -5"? closer to zero) -6 is less than -5 (it's "more negative" -- further away from zero)
The correct answer is C.
You can tell this by factoring the equation to get the zeros. To start, pull out the greatest common factor.
f(x) = x^4 + x^3 - 2x^2
Since each term has at least x^2, we can factor it out.
f(x) = x^2(x^2 + x - 2)
Now we can factor the inside by looking for factors of the constant, which is 2, that add up to the coefficient of x. 2 and -1 both add up to 1 and multiply to -2. So, we place these two numbers in parenthesis with an x.
f(x) = x^2(x + 2)(x - 1)
Now we can also separate the x^2 into 2 x's.
f(x) = (x)(x)(x + 2)(x - 1)
To find the zeros, we need to set them all equal to 0
x = 0
x = 0
x + 2 = 0
x = -2
x - 1 = 0
x = 1
Since there are two 0's, we know the graph just touches there. Since there are 1 of the other two numbers, we know that it crosses there.
Here's a list that might help:
Odd functions: sinx, tanx, cotx, cscx
Even functions: cosx, secx
