The next system of linear equations is said to be underdetermined because there are more variables than equations.
x1+2x2-3x3 = 4
2x1-x2+4x3 = -3
Similarly, the following system is overdetermined because there are more equations than variables.
x1+3x2 = 5
2x1-2x2 = -3
-x1+ 7x2 = 0
You can explore whether the number of variables and the number of equations have
any bearing on the consistency of a system of linear equations. For Exercises 1–4,
if an answer is yes, give an example. Otherwise, explain why the answer is no.
1. Can you find a consistent underdetermined linear system?
2. Can you find a consistent overdetermined linear system?
3. Can you find an inconsistent underdetermined linear system?
4. Can you find an inconsistent overdetermined linear system?
5. Explain why you would expect an overdetermined linear system to be
inconsistent. Must this always be the case?
6. Explain why you would expect an underdetermined linear system to have
an infinite number of solutions. Must this always be the case?

i think it will be zero as slope has a formula of difference between initial and final distance divided by itz time interval

the correct option is b

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.