(85 + 84 + 78 + x) / 4 = 86
(247 + x) / 4 = 86
247 + x = 86 * 4
247 + x = 344
x = 344 - 247
x = 97 <=== her score was 97
Step-by-step explanation: 29-8=21 so im guessing just subtract bc from Ac and that's your evidence maybe?
Answer:
Step-by-step explanation:
P = 3,000
r = 5% = 0.05
t = 9 years
Amount after 9 years is
A = P (1+ r)^t
A = 3,000(1+ 0.05)^9 = 3,000 * 1.05^9 = 4653.984648 ≈ $4,653.98
The balance after 9 years, rounded to the nearest cent is
$4,653 and 98 cents
The given matrix equation is,
.
Multiplying the matrices with the scalars, the given equation becomes,
![\left[\begin{array}{cc}1.5x&9\\12&6\end{array}\right] +\left[\begin{array}{cc}y&4y\\3y&2y\end{array}\right] =\left[\begin{array}{cc}z&z\\6z&2\end{array}\right] \\](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1.5x%269%5C%5C12%266%5Cend%7Barray%7D%5Cright%5D%20%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dy%264y%5C%5C3y%262y%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dz%26z%5C%5C6z%262%5Cend%7Barray%7D%5Cright%5D%20%20%5C%5C%20%20)
Adding the matrices,
![\left[\begin{array}{cc}1.5x+y&9+4y\\12+3y&6+2y\end{array}\right] =\left[\begin{array}{cc}z&z\\6z&2\end{array}\right] \\](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1.5x%2By%269%2B4y%5C%5C12%2B3y%266%2B2y%5Cend%7Barray%7D%5Cright%5D%20%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dz%26z%5C%5C6z%262%5Cend%7Barray%7D%5Cright%5D%20%20%5C%5C%20)
Matrix equality gives,

Solving the equations together,

We can see that the equations are not consistent.
There is no solution.