A new bicycle sells for $300. It is on sale for off the regular price. Select all the represent the sale price of the bicycle in dollars.

A)300•1/4

B)300•3/4

C)300•(1-1/4)

D)300- 1/4

E) 300-1/4•300

The Option B, C and E shows the sale price of the bicycle in dollars.

Option B

= 225

Option C

= 224

Option E

= 225

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Choices: B) , C) and E)

Step-by-step explanation:

Answer choice A):

$300 × = $75

* This answer choice represents the calculation of discount only and not the final selling price of the bicycle.

Answer choice B):

$300 × = $225

* This answer choice represents the calculation of the final selling price of the bicycle.

Answer choice C)

$300 × (1 - ) = $225

* This answer choice represents the calculation of the final selling price of the bicycle.

Answer choice D):

$300 - 1/4

*This answer choice shows an incorrect equation which neither represents the calculation of the discount nor the final selling price of the bicycle.

Answer choice E):

$300 - × $300

Applying the rule of PEMDAS,

We start with division; 1 ÷ 4 = 0.25

Then we go to multiplication; 0.25 × $300 = $75

Finally we do the subtraction; $300 - $75 = $225

* This answer choice represents the calculation of the final selling price of the bicycle.

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.

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]