How many terms are in the following expression? 2x + 5t -6
1 answer:
You might be interested in
Question:
Which scenario best matches the linear relationship expressed in the equation y = –14x + 1,700?
1) Kent has $1,700 in his bank account and spends $14 each week.
2) Kent has $1,700 in his bank account and deposits $14 each week.
3) Kent had $1,700 in his bank account and deposited another $14.
4) Kent has $14 in his bank account and spent $1,700.
Answer:
Option 1) Kent has $1,700 in his bank account and spends $14 each week.
Step-by-step explanation:
Let x denote the number of weeks.
Option 1: Kent has $1,700 in his bank account and -14x as the negative sign denotes how much amount he spend each week.
Thus, Option 1 is the correct answer.
Option 2: Kent has $1,700 in his bank account and deposits $14 each week.
Writing it as equation, we have, which is not the expressed linear equation.
Hence, Option 2 is not the correct answer.
Option 3: Kent had $1,700 in his bank account and deposited another $14.
Writing it as equation, we have, which is not the expressed linear equation.
Hence, Option 3 is not the correct answer.
Option 4: Kent has $14 in his bank account and spent $1,700.
Writing it as equation, we have, which is not the expressed linear equation.
Hence, Option 4 is not the correct answer.
Thus, the scenario that matches the linear equation relationship expressed in the equation is Option 1.
The answer is Kent has $1,700 in his bank account and spends $14 each week.
Answer:
Step-by-step explanation:
<u>Linear Combination Of Vectors </u>
One vector is a linear combination of
and
if there are two scalars
such as
In our case, all the vectors are given in but there are only two possible components for the linear combination. This indicates that only two conditions can be used to determine both scalars, and the other condition must be satisfied once the scalars are found.
We have
We set the equation
Multiplying both scalars by the vectors
Equating each coordinate, we get
Adding the first and the third equations:
Replacing in the first equation
We must test if those values make the second equation become an identity
The second equation complies with the values of and
, so the solution is