Find the charge on the capacitor in an LRC-series circuit at t = 0.02 s when L = 0.05 h, R = 3 Ω, C = 0.02 f, E(t) = 0 V, q(0) = 7 C, and i(0) = 0 A. (Round your answer to four decimal places.) 6.39 Correct: Your answer is correct. C Determine the first time at which the charge on the capacitor is equal to zero. (Round your answer to four decimal places.) .2 Incorrect: Your answer is incorrect. s

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

mark brainliest!

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.

hi there!

the answer is: 2 and -2

step-by-step explanation:

case 1: an absolute value of positive number will still be a positive number

so absolute value of 2 is 2

case 2: an absolute value of negative number will be positive number so absolute value of -2 is 2