Unit Circle

What is the domain of g(x) = square root of X square + 2X - 8???

(1) {x|x greater than or equal to 4 or X is less than or equal to -}
(2) {x|x greater than or equal to 2 or X is less than or equal to -4}

g(x)
= sqrt(x^2 + 2x - 8)
= sqrt((x + 4)(x - 2))

For g(x) to be real, x^2 + 2x - 8 must be >= 0
Thus, (x + 4)(x - 2) >= 0
We have x<= -4 or x>= 2

If you want to check, you can substitute a number less than -4, a number between -4 and 2, as well as a number greater than 2 into x^2 + 2x - 8

Hence, the answer is (2).

How do i solve this trigonometry question which involves Sigma?

Barron's Regents book is not as specific in explaining how to solve several problems.

The Sigma's bottom notation is (i=1), notation above is 8. So it's Sigma E, above is 8, bottom is i = 1 and then (3i - i^2).

I know it's -96, but have no clue how it came to be -96.



ANSWER:
Here's how you do it step by step. :)

Sigma always denotes SUM. Therefore, you have to add all the values of the expression by substituting i = 1 to 8.

For,

i = 1: 3(1) - (1)^2 = 3 -1 = 2
i = 2: 3(2) - (2)^2 = 6 - 4 = 2
i = 3: 3(3) - (3)^2 = 9 - 9 = 0
i = 4: 3(4) - (4)^2 = 12 - 16 = -4
i = 5: 3(5) - (5)^2 = 15 - 25 = -10
i = 6: 3(6) - (6)^2 = 18 - 36 = -18
i = 7: 3(7) - (7)^2 = 21 - 49 = -28
i = 8: 3(8) - (8)^2 = 24 - 64 = -40

Therefore, the sum is: 2 + 2 + 0 + (-4) + (-10) + (-18) + (-28) + (-40) = -96.

Hope this helps.

The roots of the equation 3x^2 - 2x = 6 are?

(1) real, irrational, and unequal
(2) real, rational, and equal
(3) real, rational, and unequal
(4) imaginary

The answer is (3)

If logb2 = r and logb3 = s, then express logb9 square root of 2 in terms of r and s. The correct answer is 2s + 1/2 r but how is the answer that?

r
= logb (2)
= logb (sqrt(2))^2
= 2*logb (sqrt(2))

Thus, logb (sqrt(2)) = r/2

logb 9*(sqrt(2))
= logb 3*3*(sqrt(2))
= logb (3) + logb (3) + logb (sqrt(2))
= s + s + r/2
= 2s + r/2

Note that:
1) log a^b = b*log a for a > 0
2) log a*b = log a + log b for a,b > 0