I NEED HEL PLEASEEE Explain how to solve an equation of the form x^p/q = a for nonzero integers x, p, q, and a. What is x in terms of a, p, and q?

Drag and drop the answers into the boxes to complete this informal argument explaining how to derive the formula for the volume of a cone. since the volume of a cone is part of the volume of a cylinder with the same base and height, find the volume of a cylinder first. the base of a cylinder is a circle. the area of the base of a cylinder is , where r represents the radius. the volume of a cylinder can be described as slices of the base stacked upon each other. so, the volume of the cylinder can be found by multiplying the area of the circle by the height h of the cylinder. the volume of a cone is of the volume of a cylinder. therefore, the formula for the volume of a cone is 1/3 1/2 1/3πr^2h 1/2πr^2h πr^2h πr^2

Which set of reflections would carry trapezoid abcd onto itself? a) x-axis, y=x, y-axis, x-axis b) x-axis, y-axis, x-axis c) y=x, x-axis, x-axis d) y-axis, x-axis, y-axis, x-axis

The table shows population statistics for the ages of best actor and best supporting actor winners at an awards ceremony. the distributions of the ages are approximately bell-shaped. compare the z-scores for the actors in the following situation. best actor best supporting actor muequals42.0 muequals49.0 sigmaequals7.3 sigmaequals15 in a particular year, the best actor was 59 years old and the best supporting actor was 45 years old. determine the z-scores for each. best actor: z equals best supporting actor: z equals (round to two decimal places as needed.) interpret the z-scores. the best actor was (more than 2 standard deviations above more than 1 standard deviation above less than 1 standard deviation above less than 2 standard deviations below) the mean, which (is not, is) unusual. the best supporting actor was (less than 1 standard deviation below more than 1 standard deviation above more than 2 standard deviations below more than 1 standard deviation below) the mean, which (is is not) unusual.