geno3141

x(x + 7.5) > 38

x² + 7.5x - 38 > 0

First, find the solutions of  x² + 7.5x - 38 = 0

I used the quadratic formula and got  x =  3.465  and  x  =  -10.965

These two numbers divide the x-axis into three groups.

So, there are three groups  (x <  -10.965)        (-10.965 < x < 3.465)       (x > 3.465)

Try any number from the first group; I'll choose -20; and put it back into the original inequality:

       x(x + 7.5) > 38     ===>     -20(-20 + 7.5)  =  250 and that's > 38 go this section works.

Try any number from the second group; I'll choose 0; and put it back into the original inequality:

       x(x + 7.5) > 38     ===>     0(0 + 7.5)  =  0 and that's not > 38 go this section doesn't works.

Try any number from the third group; I'll choose 10; and put it back into the original inequality:

       x(x + 7.5) > 38     ===>     10(10 + 7.5)  =  175 and that's > 38 go this section works.

Since the first group and the third group both work, the answer is:  x < -10.965  or  x > 3.465.

For the product of two numbers to be less than or equal to zero:

1) either one can be zero; thus,  s = -4  or  s = 3.

2) the first one is negative and the second one is positive:  

              x + 4 < 0  and  s - 3 > 0

 so:         x  < -4  and  x  > 3     (but this situation isn't possible)

3) the first one is positive and the second one is negative:

              x + 4 > 0  and  s - 3 < 0

 so:         x  > -4  and  x  < 3        (this situation is possible)

Thus, the answer (from part 1 and part 3):  -4 ≤ x ≤ 3

Part a:

Domain:  set of all possible real numbers:  -∞ < x < ∞

Range:    the set of all real numbers less than or equal to 68:  -∞ < x < 68

Part b:

Domain: 0 ≤ x ≤ 4

Range:   0 ≤ y ≤ 68

Part c:

The graph is a model of the time that the ball is in the air and the height of the ball.

x represents time; it can't be less than 0, nor can it be greater than the time that the ball returns to earth (the bouncing after that is not part of this equation).

y represents height; it can't be less than 0, for the ball is not kicked out of a hole nor is it kicked into a hole.

These restrictions weren't put into the equation.

I'm not sure which way you mean the problem:

-(-1)³ · (2-4)² - (-100)(4)  =  -(-1) · (-2)² - (-400)  =  1·4 + 400  =  4 + 400  =  404

-(-1)³ · (2-4)² - (-100)^4  =   -(-1) · (-2)² - (100 000 000)  = 1 · 4 - 100 000 000  =  -99 999 996

Or, did you mean something else?

Because cos has a period of 2π; therefore, ...   cos(4π)  =  cos(2π)  =  cos(0)  =  cos(-2π)  =  cos(-4π)  ...

k^-10  =  1/k^10

So:  9k^-10 / (9k^10)  =  9 / 9k^10k^10)  =  1 / k^20.           (the 9's cancel.) 

Of that $18.00 per hour, how much do you have in take-home pay? 

How many hours do you work per week?

Do you get overtime pay?

How much of that do you spend?

If your take-home pay were $18.00 per hour and you saved all of it, you could have $50,000.00 after 2800 hours  ($50,000.00 ÷ $18.00/hour)  but this is not realistic.

If you mean 48²:     2304

If you mean √48:    √16 · √3  =  4√3 

If you have a negative exponent in the numerator, write it in the denominator with a positive exponent.

Example:   x^-7  =  1 / x^7

Example:  3x^-4 / 14  =  3 / (14x^4)

Example:  3x^-2y^5z^-4  =  3y^5 / (x^2z^4)

If you have a negative exponent in the denominator, write it in the numerator with a positive exponent.

Example:  1 / x^-8  =  x^8

Example:  3x^3 / 2y^-6  =  3x^3Y^6 / 2

Example:  4x^5y^-2Z^-5 / (3x^-3Y^6z^-3)  =  4x^5x^3z^3 / (3y^2y^6z^5)  =  4x^8 / (3y^8z^2)

2/3 is a repeating decimal because it is 0.6666666...     (the sixes never stop)

1/8 is a terminating decimal because it is 0.125     (it stops)