John and Denise work at a factory. John arrived at work before Denise and began making shoes. John had already made 20 shoes when Denise began her work. John was producing shoes at a rate of 2 shoes per hour. Denise was able to produce shoes at a rate of 3 shoes per hour. At some point, John and Denise will have produced the same number of shoes. Part A: Write a system of equation to represent the situation (the means 2 equations). Let x=hours and y=shoes. Part B: How much time passes before John and Denise produce the same number of shoes and how many shoes was that?

the answer is 1 becuz if it were 1.0 that would rip apart the entire so ya. : )

~potato.

copyright potato 2019.

e^-kt = n/n0 = 0.79  

-kt = ln 0.79  

t = - ln 0.79/0.0001  

= 2357 years

step-by-step explanation:

hello from mrbilldoesmath!

answer:   x^2 + y^2 = 1024

discussion:

the equation of a circle with center (h,k) and radius "r" is

(x-h)^2 + (y-k)^2 = r^2.

in our case

(x-0)^2 + (y-0) ^2 = 32^2 = 1024

you,

mrb

the equation of the parallel line would be y = 2x + 9

step-by-step explanation:

to start we need to find the slope of the original line. we can do that by looking at the coefficient of x, which is 2

since the slope of the original line is 2 and parallel lines have the same slope, we know the new line will also have 2 slope. now we can use that along with the point given in point-slope form to get the find equation.

y - y1 = m(x - x1)

y - 5 = 2(x + 2)

y - 5 = 2x + 4

y = 2x + 9