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 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
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...
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?