When 3 Triangle sides are known:
The longest side must be less than the sum of the other 2 sides.
5, 12 and 13 - can be constructed
2, 11 and 15 - can be constructed
3, 7 and 11 - CAN'T be constructed - the longest side is greater than the sum of the other two.
4, 8 and 15 - CAN'T be constructed - the longest side is greater than the sum of the other two
Source:
http://www.1728.org/trianinq.htm

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7 0

3 years ago

Read 2 more answers

Solve the inequality <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B2x%7D%7Bx%20%2B%203%7D%20%20%5Cleqslant%203" id="Tex

Fofino [41]

$Domain:x\neq-3\\\\\dfrac{2x}{x+3}\leq3\qquad\text{subtract 3 from both sides}\\\\\dfrac{2x}{x+3}-3\leq0\\\\\dfrac{2x}{x+3}-\dfrac{3(x+3)}{x+3}\leq0\\\\\dfrac{2x}{x+3}-\dfrac{3x+9}{x+3}\leq0\\\\\dfrac{2x-(3x+9)}{x+3}\leq0\\\\\dfrac{2x-3x-9}{x+3}\leq0\\\\\dfrac{-x-9}{x+3}\leq0\Rightarrow(x+3)(-x-9)\leq0\\\\x=-3,\ x=-9\\\\x\in(-\infty,\ -9]\ \cup\ (-3,\ \infty)$

3 0

3 years ago

How to write 10-3 two ways

nydimaria [60]

12-5=7
13-6=7
or if you were doing fractions
ten thirds
3
-
10

5 0

3 years ago