The answer should be to 1 decimal place

addition

step-by-step explanation:

the answer would be d.

step-by-step explanation:

move all variables to the left side and all constants to the right side

(x−3)^2+(y−1)^2=16

this is the form of a circle. use this form to determine the center and radius of the circle.

( x − h ) 2 + ( y − k ) 2 = r 2

match the values in this circle to those of the standard form. the variable   r   represents the radius of the circle,   h   represents the x-offset from the origin, and   k   represents the y-offset from origin.

r = 4

h = 3

k = 1

the center of the circle is found at   ( h , k ) .

center:   ( 3 , 1 )

these values represent the important values for graphing and analyzing a circle.

center:   ( 3 , 1 ).

radius:   4

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.

answer: 60%

explanation:

÷

= 60%