11. Determine the solutions to the inequality in interval notation. (x - 1)(x + 3) < 0
a. (-0, -3) U (1,0)
b. (-3,1)
c. (-0, -1) U (3)
d. [-3,1]

9514 1404 393

  b.  (-3, 1)

Step-by-step explanation:

Quadratic (or any higher-degree) inequalities are nicely solved from their factored form. The factors tell you where the function is zero. Their signs tell you whether the product is greater than or less than zero.

Here the factor (x-1) tells you the function is zero at x=1, where this factor is zero. For values of x > 1, this factor is positive.

The  factor (x+3) tells you the function is zero at x=-3, where this factor is zero. For values of x > -3, this factor is positive.

Now we know the signs of the factors are the same for x < -3 (both negative) and for x > 1 (both positive). Between these values, the factors have different signs, so the function value is negative. This interval (-3, 1) is the interval that is the solution to ...

  (x -1)(x +3) < 0

$20.01

step-by-step explanation:

the sale price is (1.00 - 0.77)($87), or 0.23($87), or $20.01.

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !