Question

SOLVE FOR X TO THE NEAREST TENTH

Answer 1

Answer:

x = 8.8

Step-by-step explanation:

First, you have to solve the side length for the smaller triangle.

$c= \sqrt{ a^{2} +b^{2} }\\c= \sqrt{5^{2}+4^{2} } \\\\c=\sqrt{25+16} \\c=\sqrt{41} \\c= 6.4$

Now for the bigger square.

$c= \sqrt{ a^{2} +b^{2} \\} \\c = \sqrt{6.4^{2} +6^{2} } \\c= \sqrt{40.96+36} \\c= \sqrt{ 76.96}\\c= 8.77$

Rounded to the nearest tenth = 8.8

https://g.co/kgs/dg245M

I use this for my pythagorean theorem questions