Remember that a quadratic equation is a parabola. The equation is of the type y = Ax^2 + Bx + C
A linear equation is a straight line. The equation is of the type y = MX + N
The soluction of that system is Ax^2 + Bx + C = MX + N
=> Ax^2 + (B-M)x + (C-N) = 0
That is a quadratic equation.
A quadratic equation may have 0, 1 or 2 real solutions. Those are all the possibilitis.
So you must select 0, 1 and 2.
You can also get to that conclusion if you draw a parabola and figure out now many point of it you can intersect with a straight line.
You will realize that depending of the straight line position it can intersect the parabola in none point, or one point or two points.
Answer:
x=5
Step-by-step explanation:
19x and its vertical angle are equal. Lets call that angle y
<y and 17x are same side interior angles. Same side interior angles add to 180 if the lines are parallel. Since l is parallel to m.
<y + 17x=180
We know that y and 19x are equal since they are vertical angles, so we can substitute this in.
19x+ 17x = 180
Combine like terms.
36x = 180
Divide each side by 36
36x/36 = 180/36
x = 5
Answer: 
Step-by-step explanation:
Given : Distance traveled by Lily= 354 miles
Time= 6 hours
Then,

Hence, the required speed of Lily = 
Answer:
The probability of the system being down in the next hour of operation is 0.3.
Step-by-step explanation:
We have a transition matrix from one period to the next (one hour) that can be written as:
![T=\left[\begin{array}{ccc}&R&D\\R&0.7&0.3\\D&0.2&0.8\end{array}\right]](https://tex.z-dn.net/?f=T%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%26R%26D%5C%5CR%260.7%260.3%5C%5CD%260.2%260.8%5Cend%7Barray%7D%5Cright%5D)
We can represent the state that system is initially running with the vector:
![S_0=\left[\begin{array}{cc}1&0\end{array}\right]](https://tex.z-dn.net/?f=S_0%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5Cend%7Barray%7D%5Cright%5D)
The probabilties of the states in the next period can be calculated using the matrix product of the actual state and the transition matrix:

That is:
![S_1=S_0\cdot T= \left[\begin{array}{cc}1&0\end{array}\right]\cdot \left[\begin{array}{cc}0.7&0.3\\0.2&0.8\end{array}\right]= \left[\begin{array}{cc}0.7&0.3\end{array}\right]](https://tex.z-dn.net/?f=S_1%3DS_0%5Ccdot%20T%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5Cend%7Barray%7D%5Cright%5D%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.7%260.3%5C%5C0.2%260.8%5Cend%7Barray%7D%5Cright%5D%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.7%260.3%5Cend%7Barray%7D%5Cright%5D)
With the inital state as running, we have a probabilty of 0.7 that the system will be running in the next hour and a probability of 0.3 that it will be down.
Answer:
AE = 4
Step-by-step explanation:
Use the formula
AE*EC = DE*EB
Since we don't know AE or EC, let AE = x and then EC will = 16 - x. The formula then becomes
x(16 - x) = 8(6)

Getting everything on one side and setting it equal to 0 will allow us to factor for x:

Factoring that gives you that
x = 4 and x = 12. That means that one of those sements has a length of 4 and the other has a length of 12. AE is shorter so it's 4 and EC = 12