Answer:
step 1: move the -4 to the right: x^2 -2x = 4
step 2: add 1 to each side: x^2 -2x + 1 = 4 + 1
step 3: factor the left side and simplify the right side: (x-1)^2 = 5
step 4: take the square root of each side: x - 1 = sqr root of 5
step 5: move the 1 to the other side:
and 
Answer:
vertex = (- 3, 2 )
Step-by-step explanation:
The equation of a parabola in vertex form is
y = a(x - h)² + k
where (h, k) are the coordinates of the vertex and a is a multiplier
y = - 4(x + 3)² + 2 ← is in vertex form
with (h, k ) = (- 3, 2 )
<span>Assuming that each number in the table corresponds to the number of dots on the upward face of the die, </span><span>the outcome of the die roll can repeat. Thus if we simulate more rolls of the die, you won;t get the same sequence since the process is at random.</span>
Guess I have corona....................
Answer:
a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)
Step-by-step explanation:
Let's solve by separating variables:

a) x’=t–sin(t), x(0)=1

Apply integral both sides:

where k is a constant due to integration. With x(0)=1, substitute:

Finally:

b) x’+2x=4; x(0)=5

Completing the integral:

Solving the operator:

Using algebra, it becomes explicit:

With x(0)=5, substitute:

Finally:

c) x’’+4x=0; x(0)=0; x’(0)=1
Let
be the solution for the equation, then:

Substituting these equations in <em>c)</em>

This becomes the solution <em>m=α±βi</em> where <em>α=0</em> and <em>β=2</em>
![x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)](https://tex.z-dn.net/?f=x%3De%5E%7B%5Calpha%20t%7D%5BAsin%5Cbeta%20t%2BBcos%5Cbeta%20t%5D%5C%5C%5C%5Cx%3De%5E%7B0%7D%5BAsin%28%282%29t%29%2BBcos%28%282%29t%29%5D%5C%5C%5C%5Cx%3DAsin%28%282%29t%29%2BBcos%28%282%29t%29)
Where <em>A</em> and <em>B</em> are constants. With x(0)=0; x’(0)=1:

Finally:
