Answer:
The answer to your question is below
Step-by-step explanation:
1.-
A [tex]\sqrt{p^{7}}[/tex] = [tex]p^{7/2} = p^{6/2} p^{1/2}[/tex] = p³[tex]\sqrt{p}[/tex]
B [tex]\sqrt{\frac{n}{25}} = \frac{\sqrt{n}}{\sqrt{25}} = \frac{\sqrt{n}}{5}[/tex]
C [tex]\sqrt{\frac{y^{2}}{7}} = \frac{\sqrt{y^{2}}}{\sqrt{7}} = \frac{y}{\sqrt{7}}{\frac{\sqrt{7} }{\sqrt{7}}}[/tex] = [tex]\frac{y\sqrt{7}}{7}[/tex]
2.-
A 3x² = 48
x² = 48/3
x² = 16
x = ±[tex]\sqrt{16}[/tex]
x = ± 4
B x² - 2 = 12
x² = 12 + 2
x² = 14
x = ± [tex]\sqrt{14}[/tex]
3.- To solve this problem use the Pythagorean theorem
c² = a² + b²
c² = 3² + 3²
c² = 9 + 9
c² = 18
c = ± [tex]\sqrt{18}[/tex]
c = ± [tex]\sqrt{(2)(3^{2})}[/tex]
A c = ± 3[tex]\sqrt{2}[/tex]
B c = ± 4.24
