Line l is parallel to line e in the figure below.
2 answers:
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Answer:
the dimension of the poster = 90 cm length and 60 cm width i.e 90 cm by 60 cm.
Step-by-step explanation:
From the given question.
Let p be the length of the of the printed material
Let q be the width of the of the printed material
Therefore pq = 2400 cm ²
q =
To find the dimensions of the poster; we have:
the length of the poster to be p+30 and the width to be
The area of the printed material can now be:
=
Let differentiate with respect to p; we have
Also;
For the smallest area
p² = 3600
p =√3600
p = 60
Since p = 60 ; replace p = 60 in the expression q = to solve for q;
q =
q =
q = 40
Thus; the printed material has the length of 60 cm and the width of 40cm
the length of the poster = p+30 = 60 +30 = 90 cm
the width of the poster = =
= 40 + 20 = 60
Hence; the dimension of the poster = 90 cm length and 60 cm width i.e 90 cm by 60 cm.
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]