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Softa [21]
3 years ago
9

How do I solve 12=3 (2x-10)+6

Mathematics
2 answers:
mrs_skeptik [129]3 years ago
6 0

Subtract 6 from both sides and you get 6=3(2x−10)

Divide both sides by 3 and you get ​6/​​3​​=2x−10 and 6/3=2 so 2=2x−10

then add 10 to both sides    2+10=2x and you get 12=2x divide by 2  

x=6 then check your answer and you get 12=12





PilotLPTM [1.2K]3 years ago
3 0

Hello.

1. switch sides

3(2x-10)+6=12

2.subtract 6 from both sides

3(2x-10)+6-6=12-6

3. simplfy

3(2x-10)=6

4. divide both sides by 3

3(2x-10)/3=6/3

5. simpfly

2x-10=2

6. add 10 to both sides

2x-10+10=2+10

7 simpfly

2x=12

Answer=6

have a nice day

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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 :

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)

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Put the value of F into equations (2) and (4)

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Substituting the values of D, E, and F into (*)

U = t² = 3 [1 − 3t²] - 2 [2+t− 5t²] + [1 + 2t]

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