We have not discussed methods by which systems of first-order differential equations can be solved. Nevertheless, systems such as dx dt = −λ1x dy dt = λ1x − λ2y dz dt = λ2y (which is discussed here) can be solved with no knowledge other than how to solve a single linear first-order equation. Suppose that time is measured in days, that the decay constants are k1 = −λ1 = −0.139629 and k2 = −λ2 = −0.004891, and that x(0) = 20, y(0) = 0, z(0) = 0. Use a graphing utility to obtain the graphs of the solutions x(t), y(t), and z(t) on the same set of coordinate axes.

( ͡° ͜ʖ ͡° ) copy and   paste it

step-by-step explanation:

(f - g)(x) = 5x^3 - 3

step-by-step explanation:

rearrange these functions vertically, with like terms in separate columns:

f(x) = 5x^3 - 2

g(x) =         +1

now subtract, column by column:

(f - g)(x) = 5x^3 - 3

d. all of the above

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.