CentsLord

order of operations here. do the parentheses first to obtain 2 + 7 = 9. then, 18 / 9 = 2. 2 times 2 = 4, and then 4 + 1 = 5. that's how I'd do it

2 + 4 + 6 = 12

nice

yes, so the volume of a triangular pyramid is \(\dfrac{hs^2}{4\sqrt{3}}\) where h is the height and s is the base side length.

in your case, s = 6, and h = 6. this gives \(\boxed{18\sqrt{3}}\) as your answer

and that's pretty good

yes, so note that AOB and BOC add up to AOC.

AOB + BOC = AOC

sub in the values given in the question to obtain

3x + 4 + 8x - 28 = 108

simple one-variable equation, x = 12.

and AOB is 3x + 4, so that's 3(12) + 4 = 40 is your answer

hey, that makes cents

500 miles = 1 inch

and

3,000 miles = 500 * 6

so

3,000 miles = 1 inch * 6 = 6 inches

6 inches is your answer

hey, that makes cents

the perimeter of this shape is actually just the same as the combined circumference of two circles with a diameter of two feet.

the circumference of a circle with a diameter of two feet is just 2pi feet

multiply that by two to get 4pi feet,

and that's your answer, 4pi feet

hey, that makes cents

ok the big triangle is just 3.5(2 + 2 + 5) / 2 = 15.75 units of area

the smaller triangles are just 2 and 5, with a combined area of 7

then the square is 4

so that's 15.75 + 7 + 4 = 26.75

hey, that makes cents

yeah, you're on the right track! so, there's a 1/6 probability of rolling a 1 (and losing 3 bucks), a 1/3 probability of rolling a composite (and winning/losing 0), and a 1/6 probability of winning each of 2, 3, and 5. 

So... the expected value would just be \(\dfrac{1}{3} \cdot \$0 + \dfrac{1}{6} \cdot (\$2+\$3+\$5) + \dfrac{1}{6} \times -\$3 \approx \boxed{1.17}.\)

hey, that's pretty good!

(OP speaking here) Noooo!!! I thought I was right, but I was wrong. I don't even know what the answer is anymore :(

finition A polygon is said to be convex if it fully contains all line segments drawn between any two points of the polygon. Once four vertices have been selected, the quadrilateral is determined, regardless of the order of the vertices. So this is just a combination problem. We seek the number of combinations of 10 things taken 4 at a time: C(10, 4) = 10C4 = 10! 4! / (10 − 4)! = 10 × 9 × 8 × 7 / 4 × 3 × 2 × 1 = 210

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