what is the value of x by rounding to the nearest tenths

Answer:
x = 20.7
Step-by-step explanation:
So this is Pythagorean's Theorem
[tex]a^{2}+b^{2} = c^{2}[/tex]
[tex]}17.5^{2}+11^{2}=c^{2}[/tex] = 306.25 + 121 = [tex]c^{2}[/tex]
427.25 = [tex]c^{2}[/tex]
So we find the square root
c = 20.67003
To the nearest tenth it is 20.7
Answer:
x = 20.7
Step-by-step explanation:
1) Use the Pythagorean Theorem: [tex]a^{2} +b^{2} =c^{2}[/tex]
- C is always the hypotenuse of the triangle (The longest side in a right triangle)
2) Plug in 17.5 for A and 11 for B to make [tex]17.5^{2} +11^{2} =x^{2}[/tex]
3) You should end up with [tex]306.25 +121 =x^{2}[/tex]
4) Add the two numbers to get [tex]427.25=x^{2}[/tex]
5) Take the square root of both sides of the equation: [tex]\sqrt{427.25} =\sqrt{ x^{2} }[/tex]
6) You will end up with a number with many decimals, but if rounding to the nearest tenth (one decimal place), you will find that [tex]20.7=x[/tex]