Answer: OPTION B
Step-by-step explanation:
To solve this problem you must add the areas of each rectangle that form the prism;
The area of a rectangle can be calculated with the formula:

There are three pairs of equal rectangles, then you can find the surface area of 3 rectangles and multiply each one by 2, then you obtain that the surface area of the prism is:

500
0.87
2.9
It’s right I got it wrong two times and the program told me those were the right answers
Answer: the first one is C
the second one is B
The best way to put them into a number line is to make sure that they've all got the same denominator, which in this case, will be 24, as this is the lowest common multiple of 3,4 and 8. You will then need to multiply these fractions accordingly. The first one (1/4) will need to be multiplied by 6 to get a denominator of 24, so this will then give you a result of 6/24. The second one (2/3), will need to be multiplied by 8, which will then give you a result of 16/24. The third one (3/4)will also need to be multiplied by 6, giving you 18/24. The final one (6/8) will need to be multiplied by 3, giving you 18/24. You will then need to put these in order, which will be- 1/4, 2/3, 3/4, and 6/8.
The alternative method is to turn these into decimals- 1/4 is equivalent to 25%, which is equal to 0.25. 2/3 is equivalent to 66.66%, which is equivalent to 0.66. 3/4 is equivalent to 75%, which is equal to 0.75. 6/8 is also equivalent to 75%, which is equal to 0.75, you can then easily line them up this way.
Hope this has been able to help you
Part A:
Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4
The perimeter of the square is given by 4(x + 4) = 4x + 16
The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12
For the perimeters to be the same
4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2
The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.
Part B:
The area of the square is given by

The area of the rectangle is given by 2(3x + 4) = 6x + 8
For the areas to be the same

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
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