Asked by A
To the nearest tenth of an inch, find the lengths of a diagonal of a square whose side lengths are 8 inches.
a. 8
b. 9.5
c. 11.3
d. 16
Can someone explain this to me I don't understand the question?
a. 8
b. 9.5
c. 11.3
d. 16
Can someone explain this to me I don't understand the question?
Answers
Answered by
I do connexus
The answer is C
Answered by
A
thank you
Answered by
I do connexus
Your welcome also sorry I didn't have an explanation for this but I hope this helped :)
Answered by
mathhelper
Just giving the answer does NOT help!
If you study this topic you must be learning about Pythagoras
In this case you have a square, where the sides are equal, so
x^2 + x^2 = 8^2
2x^2 = 64
x^2 = 32
x = ....
If you study this topic you must be learning about Pythagoras
In this case you have a square, where the sides are equal, so
x^2 + x^2 = 8^2
2x^2 = 64
x^2 = 32
x = ....
Answered by
mathhelper
oops! misread the question....
we know the sides are 8
so d^2 = 8^2 + 8^2 = 128
d = √128 = ....
we know the sides are 8
so d^2 = 8^2 + 8^2 = 128
d = √128 = ....
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