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Ivanshal [37]
3 years ago
13

The length of a rectangle is represented by the function L(x) = 2x. The width of that same rectangle is represented by the funct

ion W(x) = 8x2 − 4x + 1. Which of the following shows the area of the rectangle in terms of x?
(L + W)(x) = 8x2 − 2x + 1
(L + W)(x) = 8x2 − 6x + 1
(L ⋅ W)(x) = 16x3 − 4x + 1
(L ⋅ W)(x) = 16x3 − 8x2 + 2x
Mathematics
2 answers:
denis-greek [22]3 years ago
6 0
I believe the correct answer from the choices listed above is the last option. The area of the rectangle can be calculated by the expression given as <span>(L ⋅ W)(x) = 16x3 − 8x2 + 2x. We obtian this as follows:
</span>

Area = LW
Area = 2x (<span>8x^2 − 4x + 1)
Area = 16x^3 - 8x^2 + 2x</span>
Dahasolnce [82]3 years ago
3 0

Answer: Area = LW

Area = 2x (8x^2 − 4x + 1)

Area = 16x^3 - 8x^2 + 2x

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sweaters already on sale for 20% off the regular price cost $28 when purchased with a promotion coupon that allows an additional
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The original price was $50.

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JUST ANSWER PLEASE!!! QUICK
jeka57 [31]

Answer:

<u>Options 1 and 3</u>

Step-by-step explanation:

We should know that, the system of linear equations can be treated as matrices, i.e: we can modify or make any operations provide that we must apply the same operation for all terms of each equation.

Given:  the solution for the following system is (2,9)

Px + Qy = R  ⇒(1)

Tx + Uy = V  ⇒(2)

We will check which system of equation has the same solution.

<u>System A)</u>  Px + Qy = R

                  (P+T)x + (Q+U)y = R+V  ⇒(3)

So, By summing (1) and (2) we will get the equation (3)

So, system A has the same solution (2,9)

<u>System B)</u> Px + Qy = R

                 (P+2T)x + (Q+2U)y = R-2V  ⇒(4)

By multiplying equation (2) by 2 and add with equation (1), we will get:

 (P+2T)x + (Q+2U)y = R+2V

Which is not the same as equation (4)

So, system B has not the same solution (2,9)

<u>System C)</u> (T-P)x + (U-Q)y = V-R  ⇒(5)

                  Tx + Uy = V  

By multiplying equation (1) by -1 and add with equation (2), we will get the equation (5)

So, system C has the same solution of (2,9)

<u>System D)</u> (T-P)x + (Q+U)y = V-R  ⇒(6)

                  Tx + Uy = V  

We cannot get equation (6) by the same operation over equation (1)

Note the coefficient of x and y⇒ (T-P) and (Q+U)

They must be (T+P) and (Q+U) <u>OR </u>(T-P) and (Q-U)

So, system D has not the same solution of (2,9)

<u>System E)</u> (5T-P)x + (5U-Q)y = V-5R ⇒ (6)

                  Tx + Uy = V  

By subtract equation (1) from 5 times equation (2), we will get:

(5T-P)x + (5U-Q)y = 5V-R

Which is not the same as equation (6)

So, system E has not the same solution (2,9)

As a conclusion, the systems which have the same solution are:

<u>Options 1 and 3</u>

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