The range is all positive values...
R{y | y>0}
Range > "Y" such that y is greater than 0
Answer:
= 11y + 1z
Step-by-step explanation:
The answer is zero, i think
Answer:
<h3>
x = 2</h3>
Step-by-step explanation:
use Pythagorean theorem:
a² + b² = c²
where a = x
b = 8/2 = 4
c = √20
plugin values into the formula:
x² + 4² = (√20)²
x² + 16 = 20
x² = 20 - 16
x = √4
x = 2
5^-2=1/5^2=1/25
2^-4=1/2^4=1/16
5^3=5*5*5=125
2^5=2*2*2*2*2=32
Hope this helped :)
