\(\frac{1000000}{\frac{100000+150000}{2}} = \frac{1000000 \cdot 2}{250000} = \boxed{8}\)
As a hint, the sum of angles in a triangle is 180 degrees.
\(\sqrt{\pi x^4} = \boxed{x^2 \sqrt{\pi}}\)
\(20+40 = \boxed{60}\)
Let \(z = a + bi \)
\((a^2-b^2) + 2abi + 2a + 2bi = 31 - 8i\)
Separate real and imaginary components.
\(a^2-b^2 + 2a = 31\)
\(2ab+2b=-8\)
The base is 24 centimeters, so the legs are 28 centimeters each. So, the perimeter is 24 + 2(28) = 80 centimeters.
The height of the triangle is \(\sqrt{28^2 - 12^2} = 8\sqrt{10}.\) That makes the area \(\frac{1}{2}(8\sqrt{10})(24) = \boxed{96\sqrt{10}}\) cm^2.
Let n be the number of students. We know that
n = 2 (mod 4)
n = 1 (mod 5)
15 students are girls, so there is a maximum of 29 students.
One solution for n is 26. So, the teacher has 26 students.
We have:
Ad + 2B = 4d + 6
Match the corresponding spots.
A = 4
B = 3
Put simply, multiply by the inverse.
\(\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c} = \frac{ad}{bc}\)
W = Fd
= (40 N)(0.8 m/s * 1800 s)
= 57600 J
