Let us determine the product of 438 and 17 by partial products.
Consider 
and 
So, 
= 
= 7446
Therefore, the product of 438 and 17 is 7446.
No, Sandra's answer is not correct.
Because she should have expressed 17 as (10+7), then if she multiplied (438 by 10) and (438 by 7). And, then added the results.Then her answer would be correct.
Answer:
NO IT IS NOT
Step-by-step explanation:
16 1/2 = 16.5 1/28 = 0.0357
NOT THE SAME!!!!!
Please mark brainiest if this is helpful! :D
Answer:
If k = −1 then the system has no solutions.
If k = 2 then the system has infinitely many solutions.
The system cannot have unique solution.
Step-by-step explanation:
We have the following system of equations
The augmented matrix is
![\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-2%263%262%5C%5C1%261%261%26k%5C%5C2%26-1%264%26k%5E2%5Cend%7Barray%7D%5Cright%5D)
The reduction of this matrix to row-echelon form is outlined below.
![\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-2%263%262%5C%5C0%263%26-2%26k-2%5C%5C2%26-1%264%26k%5E2%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-2%263%262%5C%5C0%263%26-2%26k-2%5C%5C0%263%26-2%26k%5E2-4%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-2%263%262%5C%5C0%263%26-2%26k-2%5C%5C0%260%260%26k%5E2-k-2%5Cend%7Barray%7D%5Cright%5D)
The last row determines, if there are solutions or not. To be consistent, we must have k such that
Case k = −1:
![\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%26-2%263%262%5C%5C0%263%26-2%26-1-2%5C%5C0%260%260%26%28-1%29%5E2-%28-1%29-2%5Cend%7Barray%7D%5Cright%5D%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%26-2%263%262%5C%5C0%263%26-2%26-3%5C%5C0%260%260%26-2%5Cend%7Barray%7D%5Cright%5D)
If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.
Case k = 2:
![\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%26-2%263%262%5C%5C0%263%26-2%262-2%5C%5C0%260%260%26%282%29%5E2-%282%29-2%5Cend%7Barray%7D%5Cright%5D%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%26-2%263%262%5C%5C0%263%26-2%260%5C%5C0%260%260%260%5Cend%7Barray%7D%5Cright%5D)
This gives the infinite many solution.
To find the 20th term in this sequence, we can simply keep on adding the common difference all the way until we get up to the 20th term.
The common difference is the number that we are adding or subtracting to reach the next term in the sequence.
Notice that the difference between 15 and 12 is 3.
In other words, 12 + 3 = 15.
That 3 that we are adding is our common difference.
So we know that our first term is 12.
Now we can continue the sequence.
12 ⇒ <em>1st term</em>
15 ⇒ <em>2nd term</em>
18 ⇒ <em>3rd term</em>
21 ⇒ <em>4th term</em>
24 ⇒ <em>5th term</em>
27 ⇒ <em>6th term</em>
30 ⇒ <em>7th term</em>
33 ⇒ <em>8th term</em>
36 ⇒ <em>9th term</em>
39 ⇒ <em>10th term</em>
42 ⇒ <em>11th term</em>
45 ⇒ <em>12th term</em>
48 ⇒ <em>13th term</em>
51 ⇒ <em>14th term</em>
54 ⇒ <em>15th term</em>
57 ⇒ <em>16th term</em>
60 ⇒ <em>17th term</em>
63 ⇒ <em>18th term</em>
66 ⇒ <em>19th term</em>
<u>69 ⇒ </u><u><em>20th term</em></u>
<u><em></em></u>
This means that the 20th term of this arithemtic sequence is 69.
Answer:
A) x = 6
Step-by-step explanation:
4x - 2x + 6 = x + 12
Combine like terms
2x +6 = x+12
Subtract 6 from each side
2x+6-6 = x+12-6
2x = x+6
subtract x
2x-x = x+6-x
x = 6
