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soldi70 [24.7K]
2 years ago
13

What is the area of this quadrilateral?

Mathematics
2 answers:
masya89 [10]2 years ago
4 0

\bold{ANSWER:}
30 square feet (REAL ANSWER)

\bold{SOLUTION:}

timofeeve [1]2 years ago
3 0

Answer:

42 square feet

Step-by-step explanation:

10 ×3= 30

4×3=12 divided by two =6 ×2= 12

(multiply because there are two triangular ends)

12+30=42

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30% of 500 round to the nearest hundredth
Yanka [14]

Answer: 30% of 500 is 150.

7 0
3 years ago
A triangle is removed from the rectangle. Find the area of the remaining figure.
Mariulka [41]

Answer

C) 160 in^2


Step by step explanation

First, let's find the area of the rectangle.

Area of a rectangle = length x width

The area of the rectangle = 20 * 10

= 200 in^2

Now let's find the area of the triangle that is removed.

Area of a triangle = 1/2 base * height.

Here base = 10 and height = 8. Plug in these values into the formula, we get

Area of the triangle = 1/2 * 10*8

= 5* 8

The area of the triangle = 40 in^2

The area of the remaining figure = Area of the rectangle - area of the triangle

=  200 - 40

= 160 in^2

The answer is "160 in^2"

Thank you.

5 0
3 years ago
3(2x + 5) 2 2(x + 6)
puteri [66]

Answer:

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Step-by-step explanation:

6 0
3 years ago
Please help me to solve it​
jarptica [38.1K]

Answer:

4 is b which is equivalent to 4\5

5 0
2 years ago
14a^8c^9 + 2a^4c^3 [Algebra]
rosijanka [135]

Answer:

im not the best at algebra but

2a+3b+4c

Step-by-step explanation:

subtract

(2a−3b+4c) from the sum of (a+3b−4c),(4a−b+9c) and (−2b+3c−a).

add

=(a+3b−4c)+(4a−b+9c)+(−2b+3c−a)

=(a+4a−a)+(3b−b−2b)+(−4c+9c+3c)

=4a+8c

then subtract

=(4a+8c)−(2a−3b+4c)

=4a+8c−2a+3b−4c

=2a+3b+4c

3 0
3 years ago
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