Answer:
[tex]c=1, -1[/tex]
Step-by-step explanation:
Use the Pythagorean theorem since you are working with a right triangle:
[tex]a^2+b^2=c^2[/tex]
The legs are a and b and the hypotenuse is c. The hypotenuse is always opposite the 90° angle. Insert the appropriate values:
[tex]0.8^2+0.6^2=c^2[/tex]
Solve for c. Simplify the exponents ([tex]x^2=x*x[/tex]):
[tex]0.64+0.36=c^2[/tex]
Add:
[tex]1=c^2[/tex]
Isolate c. Find the square root of both sides:
[tex]\sqrt{1}=\sqrt{c^2}\\\\\sqrt{1}=c[/tex]
Simplify [tex]\sqrt{1}[/tex]. Any root of 1 is 1:
[tex]c=[/tex] ±[tex]1[/tex] *
[tex]c=1,-1[/tex]
Finito :D
*It can be either positive of negative 1 because 1×1=1 and (-1)×(-1)=1
