I suppose
The vectors that span 
form a basis for 
if they are (1) linearly independent and (2) any vector in can be expressed as a linear combination of those vectors (i.e. they span ).
Compute the Wronskian determinant:
The determinant is non-zero, so the vectors are linearly independent. For this reason, we also know the dimension of is 3.
Write an arbitrary vector in as 
. Then the given vectors span if there is always a choice of scalars 
such that
which is equivalent to the system
The coefficient matrix is non-singular, so it has an inverse. Multiplying both sides by that inverse gives
so the vectors do span .
The vectors comprising form a basis for it because they are linearly independent.
We can use Law of Cosines to solve for the angle of Z. The solution is shown below:
cos C=(a²+b²-c²)/2ab
cos Z = (yz² + xz² - xy² )/2*yz*xz
cos Z = (20² + 25 - 13²)/2*20*25
cos Z = 856 / 1000
Z=31.13°
The answer is angle 31.13°.
Answer:
p=7x
Step-by-step explanation:
49x^[2] + 28x - 10 = p^[2] + 4p -10
This equation is in the form a^[2]x + bx + c.
<u><em>The 'c' is common for both equations, this means the 'a' and 'b' must also be common. </em></u>
There are two ways to find p: 'a' or 'b'
<u>a method</u>
49x^[2] = p^[2]
=> The square root of both sides = 7x = p
<u>b method</u>
28x = 4p
28x/4 = 4p/4
7x = p
To find perimeter you add up all the sides so the answer is 210
Answer:
4:120
Step-by-step explanation:
for 120 student they would be 4 teacher needed per 30 student
4:120
