Please HELP! I DON'T KNOW HOW CAN I SOLVE THIS
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Answer:
Step-by-step explanation:
The Pythagorean Theorem:
[tex]leg^2+leg^2=hypotenuse^2\to a^2+b^2=c^2[/tex]
[tex]41.\\\\a=1,\ b=1,\ c=x\\\\x^2=1^2+1^2\\\\x^2=1+1\\\\x^2=2\to x=\sqrt2[/tex]
[tex]42.\\a=1, b=2,\ c=x\\\\x^2=1^2+2^2\\\\x^2=1+4\\\\x^2=5\to x=\sqrt5[/tex]
[tex]43.\\a=3,\ b=x,\ c=5\\\\3^2+x^2=5^2\\\\9+x^2=25\qquad\text{subtract 9 from both sides}\\\\x^2=16\to x=\sqrt{16}\to x=4[/tex]
[tex]44.\\\\a=4,\ b=x,\ c=8\\\\4^2+x^2=8^2\\\\16+x^2=64\qquad\text{subtract 16 from both sides}\\\\x^2=48\to x=\sqrt{48}\\\\x=\sqrt{16\cdot3}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\x=\sqrt{16}\cdot\sqrt3\\\\x=4\sqrt3[/tex]
Answer:
See below.
Step-by-step explanation:
These are exercises in the application of the Pythagoras theorem:
The square on the hypotenuse = the sum of the squares on the other 2 sides.
So For No. 42:
x^2 = 121 + 2^2
x^2 = 5
x = √5 .
No. 41.
x^2 = 1^1 + 1^2
x^2 = 2
x = √2.
No 43.
5^2 = 3^2 + x^2
x^2 = 5^2 - 3^2
x^2 = 25 - 9 = 16
x = 4.