The answer for your question is 1,950
Answer:
Step-by-step explanation:
the first one
Answer:
Step-by-step explanation:
y = sin(t^2)
y' = 2tcos(t^2)
y'' = 2cos(t^2) - 4t^2sin(t^2)
so the equation become
2cos(t^2) - 4t^2sin(t^2) + p(t)(2tcos(t^2)) + q(t)sin(t^2) = 0
when t=0, above eqution is 2. That is, there does not exist the solution. so y can not be a solution on I containing t=0.
A line that is parallel has the same slop where a line that is perpendicular has a slop that is negative and reciprocal.
so for the parallel one you don't need to worry about the slop because it will be 2/3x. But yous the point slope equation form
y-y1=m(x-x1)
y+5=2/3(x+2)
y+5=2/3x+ 4/3
y=2/3x-11/3
-2/3x+y=-11/3
multiple by -1 so A inst negative
2/3-y=11/3
For a line that is perpendicular you just need to flip the original 2/3x slope and make it negative.
y+5=-3/2(x+2)
y+5=-3/2x-3
y=-3/2x-8
3/2x+y=-8
Answer:
Infinite amount of solutions.
General Formulas and Concepts:
- Order of Operations: BPEMDAS
- Regular + Equality Properties
Step-by-step explanation:
<u>Step 1: Define equation</u>
6y + 4 - 3y - 7 = 3(y - 1)
<u>Step 2: Solve for </u><em><u>y</u></em>
- Combine like terms: 3y - 3 = 3(y - 1)
- Distribute 3: 3y - 3 = 3y - 3
- Subtract 3y on both sides: -3 = -3
Here we see that there will be infinite amount of solutions. We can plug in any number <em>y</em> and it will render the equation true.
