The longest possible integer length of the third side of the triangle is 6 < x < 28
The sum of any two sides must be greater than the third side for a triangle to exist
let the third side be x
x + 11 > 17 and x + 17 > 11 and 11 + 17 > x
x > 6 and x > - 6 and x < 28
The longest possible integer length of the third side of the triangle is 6 < x < 28
The length of the 3 sides of a triangle needs to always be among (however no longer the same) the sum and the difference of the opposite two sides. As an example, take the instance of two, 6, and seven. and. consequently, the third side period should be extra than 4 and less than 8.
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Answer:
Theo worked on it for 12 hours. Kade worked on it for 15 hours.
Step-by-step explanation: 12 plus 15 equals 27.
Answer:
Step-by-step explanation:
So first you got to simlipy both sides of the equation.Like this:
10x-3x+2=6x+6
7x+2=6x+6
Then you have to move the variables to one side and the other nums to the other side:
7x-6x = 6 -2
Then you have to simplify.
x = 4
Hoped it helped!
Because of the measurements, no but if you notice closely teh shape looks translated or a reflection
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<h3>x²-4=x+2</h3>
<h3>=>x²-x-4-2=0</h3>
<h3>=>x²-x-6=0</h3>
<h3>=>x²-3x+2x-6=0</h3>
<h3>=>x(x-3)+2(x-3)=0</h3>
<h3>=>(x-3)(x+2)=0</h3>
<h3>=>x=3 and x= -2</h3>
<h2>iz the correct answer!</h2>
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