Find the range of possible measures for the third side of a triangle is the first two sides are 9 and 9.
1 answer:
<span>In a triangle the sum of two sides is always greater than the third side.
9-9 < x < 9+9
0 < x < 18 </span>← answer
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Answer:
e=1
Step-by-step explanation:
(e+1)×3=6
3e+3=6
3e=6-3
3e=3
e=3/3
e=1
:)
Answer:
8
Step-by-step explanation:
B^2 = C^2 -A^2 89-25 = 64 sqrt(64) is 8
Answer:
p = 9/4 or 2.25
Step-by-step explanation:
4p + 1 = 10
-1 -1
4p = 9
Divide 4 on both sides
p = 9/4 or 2.25
The real roots are 3 and -3.
Imaginary roots:-
x - (2 - i) = 0
x = 2 -i is one imaginary root and the other is 2 + i
Answer:
x + 2
Step-by-step explanation:
3x^3 - 12x
=3x(x^2 - 4)
= 3x(x + 2)(x - 2)
so (x+2), (x - 2) and 3x are the factors of 3x^3 - 12x
answer:
x + 2