X=36
Divide the corresponding side (16) by 4/7
H (x) = 5x^2+40x+64
y = (5x^2+40x+60) + 4
y = 5(x^2+8x+12) + 4
y = 5(x+2)(x+6) + 4
y-4 = 5(x+2)(x+6)
This means that the solutions, or where the parabola crosses the x-axis, are at x+2=0 and x+6=0, or x=-2, -6
So the midpoint of -2 and -6 is -4, the axis of symmetry is x=-4
Hmm, interesting
one way would be to multply it out or set it equal to 11x where x is a whole number (if x is not a whole number, then it is not divisible)
11x=7^6+7^5-7^4
undistribute 7^4
11x=(7^4)(7^2+7^1-1)
11x=(7^4)(49+7-1)
11x=(7^4)(55)
56=5*11
11x=(7^4)(5)(11)
divide by 11
x=5(7^4)
aka, find if 11 is a factor of that number
x=5(7^4)
<u>Solution-</u>
As given in △ABC,

As from the properties of trigonometry we know that, the greater the angle is, the greater is the value of its sine. i.e

According to the sine law,

In order to make the ratio same, even though m∠A>m∠B>m∠C, a must be greater than b and b must be greater than c.

Also given that its perimeter is 30. Now we have to find out whose side length is 7. So we have 3 cases.
Case-1. Length of a is 7
As a must be the greatest, so b and c must be less than 7. Which leads to a condition where its perimeter won't be 30. As no 3 numbers less than 7 can add up to 30.
Case-2. Length of b is 7
As b is greater than c, so c must 6 or less than 6. But in this case the formation of triangle is impossible. Because the triangle inequality theorem states that the sum of any 2 sides of a triangle must be greater than the measure of the third side. If b is 7 and c is 6, then a must be 17. So no 2 numbers below 7 can add up to 17.
Case-3. Length of c is 7
As this is the last case, this must be true.
Therefore, by taking the aid of process of elimination, we can deduce that side c may have length 7.