the longest side of an acute isosceles triangle 8 centimeters . Round to the nearest tenth, what is the smallest possible length
of one of two congruent sides?
2 answers:
Let x be the <span>length of each of two congruent sides.
</span>The triangle will be аcute if:
x² + x² > 8²
2x² > 64
x² > 32
x > √32
x > 5.657
So, the smallest possible length of one of two congruent sides have to be 5.7 cm (<span>to the nearest tenth)</span>
Answer:
The answer is D: (5.7cm)
just did the quiz on Edg
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-(1/4 w + 8) + 8 = 3
-1/4 w - 8 = 3 - 8 = -5
1/4w = -8 + 5 = -3
w = 4 x -3 = -12
w = -12
The answer i got was 31.4.... if that’s wrong i’m so sorry.
BD= 53 units
Good Luck! <3
Answer:
15
Step-by-step explanation:
4/5 4*3 is 12 5*3 is 15
its 1/6 i caculated it on google
