Answer: 0.63 repeating
Step-by-step explanation:
Alrighty! So, we're going to be dividing 7/11. First off, we know that 11 cannot go into 7, so we will add a decimal point to the 7. We now have 7.0 (aka 70) divided by 11.Don't forget to add the decimal on top! How many times does 11 go into seventy? Six times. So, now we put the six on top. Six times 11 equals 66, so now we're doing to subtract 66 from 70. Your answer should be 4. And because we haven't gotten our final answer yet, we will have to add another a 0 to 7.0 (now 7.00) and add the 0 down to 4 aswell (now 40) Now, how many times does 11 goes into 40? 3. So add the 3 on top. 11x3 equals 33, so now we're going to subtract 33 from 40. You should get 7. Like the last time, since we haven't gotten a final answer, we're going to add another 0 to 7.00 (now 7.000) and bring a 0 down. You should now have 70.
Just to help you save some time, if you repeat this process, you will get a repeating decimal ( 0.63636363636363636363636363...) So, your final answer is 0.63 repeating. (Add a line above the 63 to indicate its repeating)
Sorry if this is kinda complicated, I tried my best to explain it!
^_^
<u>*Let me know if you have any questions!</u>
Answer:
We have the matrix ![A=\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&8&4\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%26-4%26-4%5C%5C0%26-8%26-4%5C%5C0%268%264%5Cend%7Barray%7D%5Cright%5D)
To find the eigenvalues of A we need find the zeros of the polynomial characteristic 
Then
![p(\lambda)=det(\left[\begin{array}{ccc}-4-\lambda&-4&-4\\0&-8-\lambda&-4\\0&8&4-\lambda\end{array}\right] )\\=(-4-\lambda)det(\left[\begin{array}{cc}-8-\lambda&-4\\8&4-\lambda\end{array}\right] )\\=(-4-\lambda)((-8-\lambda)(4-\lambda)+32)\\=-\lambda^3-8\lambda^2-16\lambda](https://tex.z-dn.net/?f=p%28%5Clambda%29%3Ddet%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4-%5Clambda%26-4%26-4%5C%5C0%26-8-%5Clambda%26-4%5C%5C0%268%264-%5Clambda%5Cend%7Barray%7D%5Cright%5D%20%29%5C%5C%3D%28-4-%5Clambda%29det%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-8-%5Clambda%26-4%5C%5C8%264-%5Clambda%5Cend%7Barray%7D%5Cright%5D%20%29%5C%5C%3D%28-4-%5Clambda%29%28%28-8-%5Clambda%29%284-%5Clambda%29%2B32%29%5C%5C%3D-%5Clambda%5E3-8%5Clambda%5E2-16%5Clambda)
Now, we fin the zeros of 
.
Then, the eigenvalues of A are 
of multiplicity 1 and 
of multiplicity 2.
Let's find the eigenspaces of A. For : 
.Then, we use row operations to find the echelon form of the matrix
![A=\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&8&4\end{array}\right]\rightarrow\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&0&0\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%26-4%26-4%5C%5C0%26-8%26-4%5C%5C0%268%264%5Cend%7Barray%7D%5Cright%5D%5Crightarrow%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%26-4%26-4%5C%5C0%26-8%26-4%5C%5C0%260%260%5Cend%7Barray%7D%5Cright%5D)
We use backward substitution and we obtain
1.
2.
Therefore,
For 
: 
.Then, we use row operations to find the echelon form of the matrix
![A+4I_3=\left[\begin{array}{ccc}0&-4&-4\\0&-4&-4\\0&8&8\end{array}\right] \rightarrow\left[\begin{array}{ccc}0&-4&-4\\0&0&0\\0&0&0\end{array}\right]](https://tex.z-dn.net/?f=A%2B4I_3%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%26-4%26-4%5C%5C0%26-4%26-4%5C%5C0%268%268%5Cend%7Barray%7D%5Cright%5D%20%5Crightarrow%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%26-4%26-4%5C%5C0%260%260%5C%5C0%260%260%5Cend%7Barray%7D%5Cright%5D)
We use backward substitution and we obtain
1.
Then,
you can just split the figure in 2 separate figures and them add the areas:
area rec. 1:
3m×3m= 9m²
area rec. 2:
3m× (2m+3m)
= 3m×5m
=15m²
total area:
9m²+15m²= 24m²
Area of a circumference=π*r²for diameter=21.2 cm---------> r=10.6 cmA=π*10.6²--------> A=π*112.36 -------> A=352.81 cm²if 2π radians (full circumference) has an area of -----------------> 352.81 cm² 3π/5 radians-------------------------------------> XX=[(3π/5)*(352.81)]/2π---------> X=105.84 cm²
Answer:
104
Step-by-step explanation:
you have to add both numbers up
