PLEASE ANSWER QUICK (02.03 MC)
Figure EFGH on the grid below represents a trapezoidal plate at its
starting position on a rotating surface:
The plate is rotated 90° about the origin in the counterclockwise
direction. In the rotated trapezoid, what are the coordinates of the
endpoints of the side congruent to side EH?

answers-
(8,4) and (5,2)
(4,-8) and (2,-5)
(-8,-4) and (-5,-2)
(-8,-4) and (-5,-7)

answer: a video game has three difficulty levels: easy, medium, and expert. the game's scoring works in such a way that a player has to score a fixed number of points to advance to the next level, and then the scoring starts over. the total score needed to clear all three difficulty levels is 7,500. two times the number of points needed to clear the easy level is 500 more than the number of points needed to clear the medium level. twice the number of points needed to clear the medium difficulty level is 1,500 more than the number of points needed to clear the expert level. if x, y, and z represent the easy, medium, and expert levels, respectively, and a is the coefficient matrix of the system of equations modeling this situation, identify the inverse matrix, a-1, and the solution matrix, x, of the system of equations.

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answers

rauschg

rauschg · hand

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step-by-step explanation:

hello!

to begin with, we're given that on earth, a rock weighs 20 pounds, and while on jupiter, that same rock weighs 42 pounds. now, we're trying to find the weight of a rock that weighs 10.5 pounds on earth, on jupiter.

to solve this, first, set up a proportion, cross-multiply, and solve it algebraically.

setting up the proportion is the hardest and the most confusing part of the problem, but it's simple.

if on earth, the rock weighs 20 lbs, and on jupiter it weighs 42 lbs, our first part of our proportion is 20/42 lbs. the second part has to be equal to each other. this means that 10.5 has to be on the numerator of the fraction, and x, the value we're trying to find, has to be on the denominator. remember that the rock weighs 10.5 lbs on earth, and we're trying to find the weight of it on jupiter (which is x).

20/42 = 10.5/x

next, you would cross-multiply.

20x = 42(10.5)

20x = 441

now, solve it algebraically by using division.

x = 22.05 lbs

this can be rounded to 22.1 lbs.

therefore, the weight of the rock on jupiter is 22.1 lbs, the fourth choice.

a 90 degree rotation across the the origin followed by a reflection across the x-axis.

step-by-step explanation: