Combine these radicals.
1 answer:
You might be interested in
For number of students equals to zero (s=0), you have the cost amount as following:
C = 0*4.5 + 50 = 0 + 50 = 50 dollars. So the first point on a graph will be (0, 50) (first number being the s coordinate-number of students, and the second number being the c coordinate-cost)
Because 0 students are going to the trip and no one will be paying 4.5 dollars, but 50 dollars will be paid as the nonrefundable deposit (no matter if the students are going or not).
For s=1, cost will be: c = 1*4.5 + 50 = 4.5 + 50 = 54.5 dollars.
Because 1 students is going, he will be paying 4.5 dollars, and deposit of 50 dollars is always paid. So the point on a graph will be (1, 54.5).
For s=2, two students are going on the trip, cost will be:
c = 2*4.5 + 50 = 9 + 50 = 59 dollars.
The graph point will be (2, 59).
For three students s = 3, cost is:
c = 3*4.5 + 50 = 13.5 + 50 = 63.5 dollars.
Because each of the 3 students are paying 4.5 dollars and a one-time deposit is paid for the group which is going to the trip (50 dollars), so the total amount is 63.5 dollars. Point on a graph will be (3, 63.5). First coordinate of a point is 3 students (s=3) and the second coordinate is 63.5 (total cost in dollars).
For s=4 (4 students), cost is: c=4*4.5 + 50 = 18 + 50 = 68 dollars. Point with coordinates on a graph is (4, 68).
Etc. So for number of students going to the trip s=24, cost would be, as per formula c = s*4.5 + 50, c=24*4.5 + 50 = 158 dollars.
I attached a graph (.png image) for this math problem.
C = 0*4.5 + 50 = 0 + 50 = 50 dollars. So the first point on a graph will be (0, 50) (first number being the s coordinate-number of students, and the second number being the c coordinate-cost)
Because 0 students are going to the trip and no one will be paying 4.5 dollars, but 50 dollars will be paid as the nonrefundable deposit (no matter if the students are going or not).
For s=1, cost will be: c = 1*4.5 + 50 = 4.5 + 50 = 54.5 dollars.
Because 1 students is going, he will be paying 4.5 dollars, and deposit of 50 dollars is always paid. So the point on a graph will be (1, 54.5).
For s=2, two students are going on the trip, cost will be:
c = 2*4.5 + 50 = 9 + 50 = 59 dollars.
The graph point will be (2, 59).
For three students s = 3, cost is:
c = 3*4.5 + 50 = 13.5 + 50 = 63.5 dollars.
Because each of the 3 students are paying 4.5 dollars and a one-time deposit is paid for the group which is going to the trip (50 dollars), so the total amount is 63.5 dollars. Point on a graph will be (3, 63.5). First coordinate of a point is 3 students (s=3) and the second coordinate is 63.5 (total cost in dollars).
For s=4 (4 students), cost is: c=4*4.5 + 50 = 18 + 50 = 68 dollars. Point with coordinates on a graph is (4, 68).
Etc. So for number of students going to the trip s=24, cost would be, as per formula c = s*4.5 + 50, c=24*4.5 + 50 = 158 dollars.
I attached a graph (.png image) for this math problem.
Answer:
(1) No matter what's the value of ,
is never in the span of
and
.
(2) The three vectors ,
, and
are always linearly dependent for all real
.
Step-by-step explanation:
<h3>(a)</h3>If is in the span of
and
, there need to exist real
and
such that
.
Assume that such and
do exist.
In other words,
.
.
.
Rewrite as an augmented matrix and row-reduce:
.
(Add four times row one to row two and times row one to row three.)
.
Note that in row two,
- Left-hand side:
;
- Right-hand side:
.
In other words, this system is inconsistent. There's no real and
that would satisfy the condition
.
Hence .
There's no real that allows
,
to be part of the span of
and
.
If the three vectors are linearly dependent, at least one of them can be expressed as the linear combination of the other two.
Note that
. In other words,
can be written as the linear combination of the other two vectors. Additionally, since the coefficient in front of
is zero, neither the exact value of
nor the value of
will make a difference. Therefore, for all
, the three vectors
,
, and
are linearly dependent.