\(ABCDE\) forms an irregular pentagon, which can be split into \(5\) triangles, each with an area of \(\sqrt3\), so the answer is \(\color{brown}\boxed{5\sqrt3}\).
There are \(9 \times 10\times 1 = 90\) options. There are \(22\) numbers that work: \(136,176,216,256,296,336,376,416,456,496,536,576,616,656,696,736,776,816,856,896,936,976\)
Thus... the answer is \({22\over 90}= \color{brown}\boxed{11\over45}\)
Add all the equations:
\(10a+10b+10c+10d=28\)
Divide by 10
\(a + b+c+d=\color{brown}\boxed{2.8}\)
\(CBD = 360 - 108 - 47 - 141 = 67 \)
\(\color{brown}\boxed{{\angle}CBD = 67}\)
Plug 5 into the equation:
\(f(5) = 3f(5-2) - 2f(5-1)\)
\(f(5)=3f(3)-2f(4)\)
\(f(3)=3f(1) - 2f(2)\)
\(f(3)=-3 - 6\)
\(f(3)=-9\)
\(f(4)=3f(2) - 2f(3)\)
\(f(4) = 9 +18\)
\(f(4)=27\)
\(f(5)= -27- 54\)
\(\color{brown}\boxed {f(5)=-81}\)
There are \(200\) integers that are divisible by \(2\)
There are \(133\) integers that are divisible by \(3\)
We need to account for multiples of both \(2\) and \(3\)
There are \(66\) intergers that are divisble by \(6\).
So, there are \(400-200-133+66 = \color{brown}\boxed{133}\)
i might have messed up somewhere, these questions are easy to slip up on.
\(\color{brown}\boxed{-7\over9}\)
The equation can be written as \(6(j^2-j-2)\).
We can divide the equation by \(6\) to get \(j^2-j-2\).
This is: \(j^2-j-2=0\)
Add \(2\) to both sides: \(j^2-j=2\)
Rewrite: \((j-0.5)^2 = 2+0.25\)
So: \(6(j-0.5)^2 = 6j^2-6j+1.5\)
We need to subtract \(13\) to get the equation, so we have: \(6(j-0.5)^2-13\)
Thus, the answer is \({-13\over-0.5} = \color{brown}\boxed{26}\)
The distance between \(A\) and \(B\) is \(12\) across, and \(6 \) up.
Because the distance between \(B\) and \(C\) is half of the distance, \(C\) will be \(6\) units across, and \(3\) units above \(B\), meaning that \(C\) is \(\color{brown}\boxed {(20,7)}\)
I brute forced it.
\(A = 15, B = 4, C = 10, D = 6\)
So, \(\color{brown}\boxed{A=15}\)
