\(A = \frac{bh}{2} = \frac{12b}{2} = 28 \quad \rightarrow \quad b = \frac{14}{3}\)
\(P = b + 2\sqrt{b^2 + h^2} = \frac{14}{3} + 2\sqrt{(\frac{14}{3})^2 + 12^2}\)
We only care about the denominator: y^2 - 8y + 2.
The denominator can't be equal to 0, so we are looking for the roots of y^2 - 8y + 2 = 0.
From Vieta's formulas, the sum is -b/a = 8.
As a general note, for a square array with \(n\) dots on each side, the number of squares is:
\(1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}\)
The zeros of the quadratic are at x = -5 and x = -2, so the equationis \(y = (x+5)(x+2).\) Expand this to get the answer.
space diagonal = \(\sqrt{3^2+ 4^2 + 5^2} \)
edit: sniped
\(\frac{-7}{3w} = 0.35\)
\(-7 = (3w)(0.35)\)
\(w = \frac{-7}{1.05} = \boxed{\frac{20}{3}}\)
(sniped)
This is arithmetic growth, since this is a linear addition of 50 students.
ax - 2ay + 2bx - 4by - 3x + 6y
a(x - 2y) + 2b(x - 2y) - 3(x - 2y)
(a + 2b - 3)(x - 2y)
To get you started, note that g(x) = (-2+x) - f(x). Substitute in f(x) to get the answer.
Let the roots of the quadratic be x and y.
sum of roots = \(-\frac{b}{a} = -\frac{-3}{2} = \frac{3}{2}\)
product of roots = \(\frac{c}{a} = \frac{27}{2}\)
sum of squares of roots = \(x^2 + y^2 = (x+y)^2 - 2xy = \frac{9}{4} - 27 = \boxed{-24.75}\)
