If the sides of a triangle are x, x+5, 2x+5. For what values of "x" would the triangle have a perimeter of at least 72?

solve

x+ x+5 + 2x+5 ≥ 72

x+x+5+2x+5 ¡Ý 72

4x+10 ¡Ý 72

4x ¡Ý 72-10

4x ¡Ý 62 Divide with 4

x ¡Ý 62/4

x¡Ý 15.5

¡Ý is symbol greate or equal

To find the perimeter of a triangle, you add the lengths of all its sides together. In this case, the sides of the triangle are x, x + 5, and 2x + 5. So, to find the perimeter, you add these three expressions:

Perimeter = x + (x + 5) + (2x + 5)

Next, simplify the expression:

Perimeter = x + x + 5 + 2x + 5
Perimeter = 4x + 10

Now, we can set up an inequality to find the values of x for which the triangle's perimeter is at least 72:

4x + 10 ≥ 72

Subtract 10 from both sides of the inequality:

4x ≥ 62

Finally, divide both sides by 4 to solve for x:

x ≥ (62 / 4)
x ≥ 15.5

Therefore, the value of x must be equal to or greater than 15.5 in order for the triangle to have a perimeter of at least 72.