In a triangle, PR is 12 more than twice PQ and QR is two more than 4 times PQ. If all three sides of the triangle have integer lengths, what is the largest possible value of x?

x = 10

Step-by-step explanation:

In a triangle, PR is 12 more than twice PQ and If all three sides of the triangle have integer lengths, what is the largest possible value of x?

From the attached diagram, x = PQ

We have sides:

PQ, PR and QR

PR is 12 more than twice PQ

PR = 12 + 2PQ

PR = 12 + 2x

QR is two more than 4 times PQ.

QR = 2 + 4(PQ)

QR = 2 + 4x

Hence: we solve using Triangle Inequality

The triangle inequality states that the sum of the lengths of two sides of a triangle must be greater than the length of the third side.

Based on this property, if you know the lengths of two sides of a triangle and are trying to find the range of lengths of the third side, you can add the two known side lengths together and subtract the smaller one from the bigger one. The third side must be greater than the sum of the other two sides and less than their difference.

PQ > PR + QR

PQ < PR - QR

Therefore:

x + 12 + 2x > 2 + 4x

3x + 12 > 2 + 4x

12 - 2 > 4x - 3x

10 = x

step-by-step explanation:

5×5×5+2(8)÷2(2)+3(3)

15+16÷4+9

31÷13

do 31÷13 and u will get ur answer

1) is right

step-by-step explanation: