A lock has two digit numeric code, with the numbers 1 through 7 available for each digit of the code. Note that numbers can be r
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Complete Question:
In P2, find the change-of-coordinates matrix from the basis B = {1 − 3t² , 2+t− 5t² , 1 + 2t} to the standard basis C = {1, t, t²}. Then, write t² as a linear combination of the polynomials in B.
Answer:
The change of coordinate matrix is :
U = t² = 3 [1 − 3t²] - 2 [2+t− 5t²] + [1 + 2t]
Step-by-step explanation:
Let U = {D, E, F} be any vector with respect to Basis B
U = D [1 − 3t²] + E [2+t− 5t²] + F[1 + 2t]..............(*)
U = [D+2E+F]+ t[E+2F] + t²[-3D-5E]...................(**)
In Matrix form;
The change of coordinate matrix is therefore,
To find D, E, F in (**) such that U = t²
D + 2E + F = 0.................(1)
E + 2F = 0.........................(2)
-3D -5E = 1........................(3)
Substituting eqn (2) into eqn (1 )
D=3F...................................(4)
Substituting equations (2) and (4) into eqn (3)
-9F+10F=1
F = 1
Put the value of F into equations (2) and (4)
E = -2(1) = -2
D = 3(1) = 3
Substituting the values of D, E, and F into (*)
U = t² = 3 [1 − 3t²] - 2 [2+t− 5t²] + [1 + 2t]
You need to determine how much paper you need to cover the lateral side of the cylinder shown in the picture. For this, you have to calculate the surface area of the cylinder, which you can do using the following formula:
Where
A is the area
π is the number pi, for the calculations we usually use up to the first two decimal values of this number, 3.14
r is the radius
h is the height of the cylinder
The given cylinder has a height of h=15m and a diameter of d=6m
To calculate the lateral area you need to use the radius. The diameter is twice the radius, so to determine the radius of the cylinder you have to divide the diameter by 2
Now you can calculate the lateral area as follows:
Charlie will need 282.6 m² to cover the lateral side of the cylinder.
Answer:
2x - y - 3z = 0
Step-by-step explanation:
Since the set
{i, j} = {(1,0), (0,1)}
is a base in
and F is linear, then
<em>{F(1,0), F(0,1)} </em>
would be a base of the plane generated by F.
F(1,0) = a+b = (2i-2j+k)+(i+2j+k) = 3i+2k
F(0,1) = a+c = (2i-2j+k)+(2i+j+2k) = 4i-j+3k
Now, we just have to find the equation of the plane that contains the vectors 3i+2k and 4i-j+3k
We need a normal vector which is the cross product of 3i+2k and 4i-j+3k
(3i+2k)X(4i-j+3k) = 2i-j-3k
The equation of the plane whose normal vector is 2i-j-3k and contains the point (3,0,2) (the end of the vector F(1,0)) is given by
2(x-3) -1(y-0) -3(z-2) = 0
or what is the same
2x - y - 3z = 0