Two small bubbles that each have a radius of 3 centimeters collide and merge into one large bubble. The area of
the large bubble is equal to the sum of the areas of the two small
bubbles. To the nearest centimeter, what is the radius of the large
bubble?
Use 3.14 for Pi

find area of 1 small bub

πr² = 3.14 × 3²

= 28.26 cm

area of 2 small bub

28.26 × 2 = 56.52 cm

area of big bub = area of 2 small bub

= 56.52

find radius by using formula

πr² = 56.52

r² = 56.52 ÷ 3.14

r² = 18

r = √18

r = 4.24 cm

round off to nearest cm

4.24 ≈ 4 cm

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length

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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