Question

The area of a poster is 2x² + x – 6. Its width is 2x – 3.
a. Find the length.
b. If x = 2 m, find the actual dimension of a poster​

Answer 1

Answer:

,

Step-by-step explanation:

Answer in attachment.

plz folløw

Answer 2

Given :-

• The area of a poster is 2x² + x – 6. Its width is 2x – 3.

To Find :-

• a. Find the length ?
• b. If x = 2 m, find the actual distance of a poster ?

Solution :-

a).

• Let the length of the poster be I units.
• And, width of the poster = 2x - 3 units.

We know that,

• Using formula,

• Area of a rectangle = Ib units

Area of the poster = l × (2x - 3) sq.units

★ According to the question :

➻ l × (2x - 3) = 2x² + x - 6

➻ l = (2x² + x - 6)/(2x - 3)

➻ l = (2x² + 4x - 3x - 6)/(2x - 3)

➻ l = [2x(x + 2) -3 (x + 2)]/(2x - 3)

➻ l = (x + 2)(2x - 3)/(2x - 3)

➻ l = (x + 2) units

• Hence, length of the poster is x + 2 units.

b).

Given :-

• x = 2m

To Find :-

• Find the actual dimension of a poster ?

Solution :-

• Length of the poster = x + 2 units
• If x = 2 m then the length becomes

➻ 2 + 2 m

➻ 4 m

And,

• Width of the poster = 2x - 3 units
• If x = 3 then width becomes

➻ 2(2) - 3 m

➻ 4 - 3 m

➻ 1 m

• Hence, length is 4 m and width is 1 m.

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