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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The final step in solving the inequality is to divide through by -2 and flip the sign
<h3>Solving Inequalities </h3>
From the question, we are to determine the final step in solving the inequality
The given inequality and the steps for solving are
Step 1: –10 + 8x < 6x – 4
Step 2: –10 < –2x – 4
Step 3: –6 < –2x
The final step will be to divide both sides of the inequalities by -2 and then flip the sign
That is,
–6 < –2x
Divide both sides by -2 and flip the sign
3 > x
OR
x < 3
Hence, the final step in solving the inequality is to divide through by -2 and flip the sign
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Answer:
y is 1 and x is 4
Step-by-step explanation:
becasue the y intercepts at 1 and the x intercepts at 4
101, of course. What else would it be????????????????
