Recent content by Augustine Duran

plugging in C and simplifying i get ##\text{x=}####\frac 8 3####\text{+}####\frac 4 3####e^{-3t}##

x(t) = \begin{cases} 3x-8 & \text{if } x > \frac 8 3 \\ 8-3x & \text{if } x \leq \frac 8 3 \end{cases} okay so seeing that my initial conditions give me x=4 i will be using 3x-8 since 4 is larger than 8/3? ##\text{3x-8 =}####e^{-3t-3c}## ##\text{x =}####\frac 1...

Homework Statement Solve: ##\frac {dx} {dt}## ##\text{= 8-3x , x(0)=4}## The Attempt at a Solution Step 1: ##\int \frac 1 {8-3x} \, dx## = ##\int \, dt## Step 2: - ##\frac 1 3## ##\text{ln|8-3x| = t+c}## From here I am going to try to get it into explicit form Step 3...

Homework Statement A person with a near point of 100 cm , but excellent distant vision, normally wears corrective glasses. But he loses them while traveling. Fortunately, he has his old pair as a spare. If the lenses of the old pair have a power of +2.55 diopters , what is his near point...

ok thanks for all the help and being patient, i really appreciate it!

ok final question. overall it seems i multiplied it by -3, which would seem to be my k. But multiplying my determinant (27) by -3 obviously gives the wrong answer. So shouldn't it be defined as 1/k instead of just k?

so... -If one row of A is multiplied by k to produce B, then detB = k ⋅ detA so correct me if I am wrong, i will still end up with the correct determinant if i multiply that determinant by k since detB = k ⋅ detA

so dividing det(R⋅A) by det R gives me det A. what's bugging me is that, do i have apply the same operations to the identity matrix to get det R, and then divide det( R⋅A) by det R every time to find the determinant? the example in the book just did some row operations then multiplied the main...

ok so the Multiplicative Property u gave me is 2 pages ahead from where I am at, I am trying to understand an example using those theorems given to me. im aware that if you swap rows the detB = - detA

unless your trying to point out that the determinant of my given matrix shouldn't change if the determinant of the identity matrix doesn't change?

to be honest i don't know. maybe you can rephrase your question so i can better understand? all i know is the theorems in listed in the book -If a multiple of one row of A is added to another row to produce a matrix B, then detB = detA. -If two rows of A are interchanged to produce B, then...

-3?

Homework Statement \begin{vmatrix} 1 & -4 & 3 & 4 \\ 0 & -9 & 6 & 8 \\ 0 & -6 & 5 & 5 \\ 0 & 0 & -3 & 2 \end{vmatrix} so the determinant of this matrix is -9, apparently I am doing something illegal in my row operations. I want to get -6 in row 3 to be 0 so... 2R2 - 3R3 = 0 0 -3 1...

So is it correct to say that both a and b are bending away from the normal at the RIGHT side of the glass at the same angle, that ray a is just shifted down?

is the reason your measuring the refraction angle from within the glass because the two rays seem to be parallel coming out of the glass?