12 times 11 is 132.
132 times $6 is $792
Hope this helps!
Well basically you just have to chose a number of sides for the side length of the square to be and move those many places to get another vertex of the square. For example if we have (-2,3) and we choose the side lengths to be 4 units than you could move 4 places up, down, left, or right to get the other vertices for the square
Hope that helps :)
Let's represent the two numbers by x and y. Then xy=60. The smaller number here is x=y-7.
Then (y-7)y=60, or y^2 - 7y - 60 = 0. Use the quadratic formula to (1) determine whether y has real values and (2) to determine those values if they are real:
discriminant = b^2 - 4ac; here the discriminant is (-7)^2 - 4(1)(-60) = 191. Because the discriminant is positive, this equation has two real, unequal roots, which are
-(-7) + sqrt(191)
y = -------------------------
-2(1)
and
-(-7) - sqrt(191)
y = ------------------------- = 3.41 (approximately)
-2(1)
Unfortunately, this doesn't make sense, since the LCM of two numbers is generally an integer.
Try thinking this way: If the LCM is 60, then xy = 60. What would happen if x=5 and y=12? Is xy = 60? Yes. Is 5 seven less than 12? Yes.
all real numbers is the domain of the function y=x+4
Percent = 75%
base = unknown ...the base is the number that follows " of "
amount = 30
