Drawing this square and then drawing in the four radii from the center of the cirble to each of the vertices of the square results in the construction of four triangular areas whose hypotenuse is 3 sqrt(2). Draw this to verify this statement. Note that the height of each such triangular area is (3 sqrt(2))/2.
So now we have the base and height of one of the triangular sections.
The area of a triangle is A = (1/2) (base) (height). Subst. the values discussed above, A = (1/2) (3 sqrt(2) ) (3/2) sqrt(2). Show that this boils down to A = 9/2.
You could also use the fact that the area of a square is (length of one side)^2, and then take (1/4) of this area to obtain the area of ONE triangular section. Doing the problem this way, we get (1/4) (3 sqrt(2) )^2. Thus,
A = (1/4) (9 * 2) = (9/2). Same answer as before.
so all those quadratics have their parent y = x², which has the graph of a "bowl".
and depending on the coefficient, the bowl is wider or not.
the larger the coefficient, the narrower the bowl, the smaller the coefficient, the wider the bowl.
well, 1/2 is smaller then 2, thus y = (1/2)x² is wider than y = 2x².
1/2 m × 1/2 m × 1/2 m = s³
your answer should be 1/8 m³
X= 1st integer
x+2= 2nd integer
x+4= 3rd integer
Add the integers together
x + (x + 2) + (x + 4)= 279
combine like terms
3x + 6= 279
subtract 6 from both sides
3x= 273
divide both sides by 3
x= 91 first integer
Substitute x=91 to find 2nd & 3rd integers
2nd Integer
=x+2
=91+2
=93
3rd Integer
=x+4
=91+4
=95
ANSWER: The three test scores are 91, 93 and 95.
Hope this helps! :)
