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3 years ago

Eliza falls asleep at 9:35 pm and wakes up 8 hours and 45 miunets later, what time does she wake uo

Mathematics

1 answer:

alexdok [17]3 years ago

6 0

Answer: Eliza wakes up at 6:25 A.M.

Step-by-step explanation:

What I did is first used the 45 minutes to make the time 10:00 P.M. which leaves you with 8 hours and 25 minutes, then i added the 8 hours onto 10:00 P.M. which leaves you at 6:00 A.M. with 25 minutes left, then you add the 25 minutes onto 6:00 A.M. which puts it at 6:25 A.M.

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Line A passes through the points (10,6) qnd (2,15). Line B passes through the points (5,9) and (14,-1).

e-lub [12.9K]

Answer:

Line A equation = y=-9/8x+69/4

Line B equation = y=-10/9x+131/9

Step-by-step explanation:

7 0

2 years ago

Christian sold tickets to the game . Good seats were $5 each and poor seats cost $2 each. 210 people attended and paid $660. How

Sholpan [36]

People who purchased good seats will be labeled as a
people who purchased poor seats will be labeled as b
therefore...
5a + 2b = 660
a + b = 210
Now, all that has to be done is for the system to be solved.
-2(a+b=210)

5a + 2b = 660
-2a - 2a = -420

3a=240
a= 80
*3 people purchased good seats.

Now we can plug that number in

5(80) + 2b = 660
400 + 2b = 660
2b = 260
b = 130
*130 people purchased poor seats

5 0

3 years ago

The system of equations may have a unique solution, an infinite number of solutions, or no solution. Use matrices to find the ge

Leno4ka [110]

Answer:

Infinite number of solutions.

Step-by-step explanation:

We are given system of equations

$5x+4y+5z=-1$

$x+y+2z=1$

$2x+y-z=-3$

Firs we find determinant of system of equations

Let a matrix A= $\left[\begin{array}{ccc}5&4&5\\1&1&2\\2&1&-1\end{array}\right]$ and B= $\left[\begin{array}{ccc}-1\\1\\-3\end{array}\right]$

$\mid A\mid=\begin{vmatrix}5&4&5\\1&1&2\\2&1&-1\end{vmatrix}$

$\mid A\mid=5(-1-2)-4(-1-4)+5(1-2)=-15+20-5=0$

Determinant of given system of equation is zero therefore, the general solution of system of equation is many solution or no solution.

We are finding rank of matrix

Apply $R_1\rightarrow R_1-4R_2$ and $R_3\rightarrow R_3-2R_2$

$\left[\begin{array}{ccc}1&0&1\\1&1&2\\0&-1&-3\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\1\\-5\end{array}\right]$

Apply $R_2\rightarrow R_2-R_1$

$\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&-1&-3\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\6\\-5\end{array}\right]$

Apply $R_3\rightarrow R_3+R_2$

$\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&0&-2\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\6\\1\end{array}\right]$

Apply $R_3\rightarrow- \frac{1}{2}$ and $R_2\rightarrow R_2-R_3$

$\left[\begin{array}{ccc}1&0&1\\0&1&0\\0&0&1\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]$

Apply $R_1\rightarrow R_1-R_3$

$\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$ : $\left[\begin{array}{ccc}-\frac{9}{2}\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]$