Acubic shaped fish tank is 40 inches tall. one quarter of the tank is filled with water. if a piece of cubic shaped metal of 30 inches in height is placed in the fish tank, what is the water level in the fish tank?

Oh, well, one quarter is equal to 1/4, the cubic shaped fish tank is 40 inches tall, and the piece of metal is 30 inches in height, that'll leave 10 inches left of space at the top of the tank, (40 - 30 = 10), and the water level will become 4 quarters full, since the metal took up most of the space and one quarter of water is larger than 10 inches. So you're answer is 4 quarters full. ;-)
I hope I helped! =D

Oh, well, one quarter is equal to 1/4, the cubic shaped fish tank is 40 inches tall, and the piece of metal is 30 inches in height, that'll leave 10 inches left of space at the top of the tank, (40 - 30 = 10)

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12

(-14+3/2b) - (1+8/2b) = -5 -5/2b(-10+b) + (7b-5) = 8b-15(7/2b-3) - (8+6b) = -5/2b-11(5+2b) + (2b+3/2) = 4b+ 13/2