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klemol [59]
1 year ago
12

Two congruent squares and a parallelogram were used to form the figure shown. 3 meters, 6 meters, 2.8 meters, 3 meters What is t

he area of the figure in square meters?
Mathematics
1 answer:
elixir [45]1 year ago
3 0

The area of a figure is the amount of space the figure occupies

The surface area of the figure is 34.8 square meters

<h3>How to determine the area of the shape</h3>

From the question, we have:

  • The dimensions of the congruent squares are 3 m by 3 m
  • The dimensions of the parallelogram are height of 2.8 m, and base of 6 m

So, the area of the figure is:

Area = 2 * Area of square + Area of parallelogram

This gives

Area = 2 * (3m * 3m) + 2.8m * 6m

Evaluate the products

Area = 18m^2 + 16.8m^2

Evaluate the sum

Area = 34.8m^2

Hence, the surface area of the figure is 34.8 square meters

Read more about surface areas at:

brainly.com/question/24571594

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BRAINLIEST ANSWER!!
Maksim231197 [3]

Answer:

1. 30(8.75) + 11t = 400

2. 12.5 hours

Step-by-step explanation:

1. What is given is you work for 30 hours per week at a gas station for $8.75 an hour. You also work as a landscaper for $11 an hour. You want to a make a total of $400 per week.

30 hours and $8.75 an hour would be equivalent to 30(8.75)

We don’t know how many hours you work as a landscaper but you earn $11 an hour, which is equivalent to 11t

Finally, you want to earn a total of $400 a week, which means the sum equals 400

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2. 30(8.75) = 262.5

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Angelina_Jolie [31]

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3 years ago
According to DeMorgan 's theorem, the complement of W · X + Y · Z is W' + X' · Y' + Z'. Yet both functions are 1 for WXYZ = 1110
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Answer:

The parenthesis need to be kept intact while applying the DeMorgan's theorem on the original equation to find the compliment because otherwise it will introduce an error in the answer.

Step-by-step explanation:

According to DeMorgan's Theorem:

(W.X + Y.Z)'

(W.X)' . (Y.Z)'

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Note that it is important to keep the parenthesis intact while applying the DeMorgan's theorem.

For the original function:

(W . X + Y . Z)'

= (1 . 1 + 1 . 0)

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For the compliment:

(W' + X') . (Y' + Z')

=(1' + 1') . (1' + 0')

=(0 + 0) . (0 + 1)

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Both functions are not 1 for the same input if we solve while keeping the parenthesis intact because that allows us to solve the operation inside the parenthesis first and then move on to the operator outside it.

Without the parenthesis the compliment equation looks like this:

W' + X' . Y' + Z'

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0 + 0 . 0 + 1

Here, the 'AND' operation will be considered first before the 'OR', resulting in 1 as the final answer.

Therefore, it is important to keep the parenthesis intact while applying DeMorgan's Theorem on the original equation or else it would produce an erroneous result.

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