To determine which of the rectangles has a different area, the areas of the four rectangles must be calculated. The rectangle with a different area is rectangle b
The dimension of the 4 rectangles are:
a. length: 4x and width: 4
b. length: 11 and width: x
c. length: 2 and width: 8x
d. length: 16x and width: 1
The area of a rectangle is:

<u>Rectangle (a)</u>


<u>Rectangle (b)</u>


<u>Rectangle (c)</u>


<u>Rectangle (d)</u>


Rectangles (a), (c) and (d) have the same area (i.e. 16x) while rectangle (b) has 11x as its area.
Hence, the rectangle with a different area is rectangle (b).
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You would use PEMDAS, which means Parenthesis, exponents, multiplication, division, addition" and then subtraction. You would solve the problem in that order
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8. 3x-4>38
3x>38+4
3x>42
x>42/3
x>14
So the answer is A. x>14
9. It's A. Two less than four times a number is 12. Because 4x (four times) and also it says 2 less than 4 times a number...