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serg [7]
3 years ago
7

the length of the rectangle shown above is (x+9). the width of the rectangle is (2x+3). Find the area of the rectangle.

Mathematics
2 answers:
Alja [10]3 years ago
8 0

(x+9)(2x+3)=2x^2+3x+18x+27=2x^2+21x+27


leonid [27]3 years ago
4 0

Answer: Area of rectangle =2x² + 21x + 27

Step-by-step explanation:

Given :

Length of the rectangle = x + 9

Width of the rectangle = 2x + 3

We have to find area of the rectangle ?

Area of rectangle = Length × Breadth

Area of the rectangle = (x +9) \times (2x+3)

Area = x \times (2x+3) + 9 \times ( 2x+3)

Area =2x^{2} + 3\times x + 9\times 2x + 9\times 3

Area =2x^{2} + 3x + 18x + 27

Area =2x^{2} + 21x + 27

∴ Area of  rectangle = 2x^{2} +21x+ 27

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You didn't give the options but let me help out.

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Step-by-step explanation:

Subject: Re: Need the math proof for 1 + 1 = 2

The proof starts from the Peano Postulates, which define the natural

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P2. If x is in N, then its "successor" x' is in N.

P3. There is no x such that x' = 1.

P4. If x isn't 1, then there is a y in N such that y' = x.

P5. If S is a subset of N, 1 is in S, and the implication

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Then you have to define 2:

Def: 2 = 1'

2 is in N by P1, P2, and the definition of 2.

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Proof: Use the first part of the definition of + with a = b = 1.

Then 1 + 1 = 1' = 2 Q.E.D.

Note: There is an alternate formulation of the Peano Postulates which

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definition of addition to this:

Def: Let a and b be in N. If b = 0, then define a + b = a.

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You also have to define 1 = 0', and 2 = 1'. Then the proof of the

Theorem above is a little different:

Proof: Use the second part of the definition of + first:

1 + 1 = (1 + 0)'

Now use the first part of the definition of + on the sum in

parentheses: 1 + 1 = (1)' = 1' = 2 Q.E.D.

3 0
3 years ago
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