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Leto [7]
2 years ago
5

Question 9 of 25

Mathematics
2 answers:
tresset_1 [31]2 years ago
8 0

Answer:

One approach to this problem is to obtain the graph for the given equation.

We need to find every intersection those functions have with the axis 'x' and 'y'

starting with g(x)

g(x=0)=0-3, first point (0,-3) it iis the crossing point with 'x' axis

g(x)=0=x-3, second point (3,0) it iis the crossing point with 'y' axis

Lets do the same for f(x)

g(x=0)=0, this leads to the first point (0,0) it iis the crossing point with 'x' axis and also, with the 'y' axis

We dont need to find any other, since always y=x

By plotting we have the attached picture

Now you can see that g(x) differs from its parent function in that is shifted 3 units to the right, and also 3 units down.

Step-by-step explanation:

KengaRu [80]2 years ago
8 0

Answer:

If f(x) = x  and  g(x) = x - 3

then g(x) = f(x) - 3

So g(x) is a translation of f(x) by 3 units down.

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In P2, find the change-of-coordinates matrix from the basis B = {1 − 3t 2 , 2 + t − 5t 2 , 1 + 2t} to the standard basis of P2.
Serjik [45]

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 :

M = \left[\begin{array}{ccc}1&2&1\\0&1&2\\-3&-5&0\end{array}\right]

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;

\left[\begin{array}{ccc}1&2&1\\0&1&2\\-3&-5&0\end{array}\right] \left[\begin{array}{ccc}D\\E\\F\end{array}\right] = \left[\begin{array}{ccc}D+2E+F\\E+2F\\-3D-5E\end{array}\right]

The change of coordinate matrix is therefore,

M = \left[\begin{array}{ccc}1&2&1\\0&1&2\\-3&-5&0\end{array}\right]

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]

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aliya0001 [1]

We know that :

Ф  Product of the slopes of two lines perpendicular to each other should be equal to -1

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