I think it’s D
My explanation is:
That I just think it’s correct but that doesn’t mean u have to put my answer tho just saying
Answer:
![6A=8B=6\left[\begin{array}{ccc}-4&-1&-9\\-9&6&9\\-1&-1&3\end{array}\right]](https://tex.z-dn.net/?f=6A%3D8B%3D6%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%26-1%26-9%5C%5C-9%266%269%5C%5C-1%26-1%263%5Cend%7Barray%7D%5Cright%5D)
![-8\left[\begin{array}{ccc}-10&-2&-5\\9&-7&-1\\9&-4&2\end{array}\right]](https://tex.z-dn.net/?f=-8%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-10%26-2%26-5%5C%5C9%26-7%26-1%5C%5C9%26-4%262%5Cend%7Barray%7D%5Cright%5D)
![=\left[\begin{array}{ccc}-24&-6&-54\\-54&36&54\\-6&-6&18\end{array}\right]](https://tex.z-dn.net/?f=%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-24%26-6%26-54%5C%5C-54%2636%2654%5C%5C-6%26-6%2618%5Cend%7Barray%7D%5Cright%5D)
![-\left[\begin{array}{ccc}-80&-56&-40\\72&-56&-8\\72&-32&16\end{array}\right]](https://tex.z-dn.net/?f=-%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-80%26-56%26-40%5C%5C72%26-56%26-8%5C%5C72%26-32%2616%5Cend%7Barray%7D%5Cright%5D)
![=\left[\begin{array}{ccc}56&50&-14\\-120&92&62\\-78&26&2\end{array}\right]](https://tex.z-dn.net/?f=%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D56%2650%26-14%5C%5C-120%2692%2662%5C%5C-78%2626%262%5Cend%7Barray%7D%5Cright%5D)
<u>---------------------------</u>
<u>hope it helps..</u>
<u>have a great day!!</u>
Help her with the dishes
Ask her whats wrong and how you can help
Pick it up and put it in the trash
Tell him its not nice and to treat other how you want to be treated.along with telling the teacher.
Answer:
y+4=-2(x-4)
Step-by-step explanation:
y-y1=m(x-x1)
y-(-4)=-2(x-4)
y+4=-2(x-4)
Answer:
A(t) = 300 -260e^(-t/50)
Step-by-step explanation:
The rate of change of A(t) is ...
A'(t) = 6 -6/300·A(t)
Rewriting, we have ...
A'(t) +(1/50)A(t) = 6
This has solution ...
A(t) = p + qe^-(t/50)
We need to find the values of p and q. Using the differential equation, we ahve ...
A'(t) = -q/50e^-(t/50) = 6 - (p +qe^-(t/50))/50
0 = 6 -p/50
p = 300
From the initial condition, ...
A(0) = 300 +q = 40
q = -260
So, the complete solution is ...
A(t) = 300 -260e^(-t/50)
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The salt in the tank increases in exponentially decaying fashion from 40 grams to 300 grams with a time constant of 50 minutes.
