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antoniya [11.8K]
2 years ago
14

what is the perimeter of the triangle. i do not know how to start this problem. any help will be greatly appreciated :))

Mathematics
1 answer:
xz_007 [3.2K]2 years ago
7 0

Given:

Point F,G,H are midpoints of the sides of the triangle CDE.

FG=9,GH=7,CD=24

To find:

The perimeter of the triangle CDE.

Solution:

According to the triangle mid-segment theorem, the length of the mid-segment of a triangle is always half of the base of the triangle.

FG is mid-segment and DE is base. So, by using triangle mid-segment theorem, we get

\dfrac{1}{2}DE=FG

DE=2(FG)

DE=2(9)

DE=18

GH is mid-segment and CE is base. So, by using triangle mid-segment theorem, we get

\dfrac{1}{2}CE=GH

CE=2(GH)

CE=2(7)

CE=14

Now, the perimeter of the triangle CDE is:

Perimeter=CD+DE+CE

Perimeter=24+18+14

Perimeter=56

Therefore, the perimeter of the triangle CDE is 56 units.

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rusak2 [61]

Answer:

1). 183

2). 83

3). 353

4). -2652

5). 59

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7). -293

8). 79

9). 78

10). 87

Step-by-step explanation:

Considering order of operation, parenthesis must be simplified first.

The solutions to the given questions are shown below;

1). \ \ 5+(7\times (3+2)^2) + 3=8 +  (7\times (5)^2)  = 8 + (7\times 25) = 8 + 175 = 183

2). \ 16+(8+ (6+2)^2) - 5=16-5 +  (8+ (8)^2)  = 11 + (8+64) = 11 + 72 = 83

3). \ (15+ 3)^2 + ((14+6)+3^2)= (18)^2 + ((20) + 9) = 324 + (29) = 353

4). \ ((11+ 2)-(24+2)^2)\times 2^2 = (13 - 676) \times 4  = (-663)\times 4 = -2652

5). \ ((4+3)^2+ 6) -5+3^2= (49 + 6) - 5 + 9= (49+6+9) - 5 = 64-5 = 59

6). \ (20+ 2)^2 +((15+7)+2^2)= (484) + (22+ 4)= 484 + 26 =510

7). \ ((12 + 4)- (15+3)^2+ 5^2 = (16 -  324) + 25 = -318 + 25 = -293

8). \ (6^2+ (24+ 3+5^2)) - 3^2= (36 + (52)) - 9 = (88) - 9 = 79

9). \ ((10-4)^2\times 2) + 2 + 2^2= ((6)^2 \times 2) + 2 + 4 = (36\times 2)+ 6 = 72 + 6 = 78

10). \ (7^2 + (16 + 8 + 2^2)) + 3^2 = (49 + 28) + 9 = 78 + 9 = 87

3 0
2 years ago
F(1) = -3<br> f(n) = 2 · f(n − 1) +1<br> f(2)=
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8 0
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