The range is all positive values...
R{y | y>0}
Range > "Y" such that y is greater than 0
32 degrees F to celcius equals 0
32 degrees F to Kelvin equals 273.15
Answer:
2x2-3x-5=0
Step-by-step explanation:
(2x2 - 3x) - 5 = 0
For this case we have the following quadratic equation:

The solutions will be given by:

Where:

Substituting the values we have:

We have two roots:

ANswer:
