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Masja [62]
3 years ago
13

Explain why both polygons A and B are Pentagons

Mathematics
2 answers:
dybincka [34]3 years ago
7 0
They both have 5 sides
Tpy6a [65]3 years ago
3 0
Your original answer was correct...
Both polygons have five sides, and five sided polygons are called pentagons.
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(12 × 6) × 10 Which of the following is equal to the expression listed above? A. (12 × 6) + 10 B. (12 × 6) × (6 × 10) C. 12 × (6
Ymorist [56]
C. 12 x (6 x10) = 720
3 0
3 years ago
Solve 7w = 56, where w is a real number.
steposvetlana [31]

Answer:

w = 8

Step-by-step explanation:

7w = 56

Divide both sides by 7

w = 56/7

w = 8

4 0
3 years ago
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lidiya [134]
Your answer is point g and h

6 0
3 years ago
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This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive an
sleet_krkn [62]

Answer:

-6re−r [sin(6θ) - cos(6θ)]

Step-by-step explanation:

the Jacobian is ∂(x, y) /∂(r, θ) = δx/δθ × δy/δr - δx/δr × δy/δθ

x = e−r sin(6θ), y = er cos(6θ)

δx/δθ = -6rcos(6θ)e−r sin(6θ), δx/δr = -sin(6θ)e−r sin(6θ)

δy/δθ = -6rsin(6θ)er cos(6θ), δy/δr = cos(6θ)er cos(6θ)

∂(x, y) /∂(r, θ) =  δx/δθ × δy/δr - δx/δr × δy/δθ

= -6rcos(6θ)e−r sin(6θ) × cos(6θ)er cos(6θ) - [-sin(6θ)e−r sin(6θ) × -6rsin(6θ)er cos(6θ)]

= -6rcos²(6θ)e−r (sin(6θ) - cos(6θ)) - 6rsin²(6θ)e−r (sin(6θ) - cos(6θ))

= -6re−r (sin(6θ) - cos(6θ)) [cos²(6θ) + sin²(6θ)]

= -6re−r [sin(6θ) - cos(6θ)]     since  [cos²(6θ) + sin²(6θ)] = 1

6 0
3 years ago
What is the correct factorization of x^2 - 2x - 15?
iragen [17]

Answer:

option b

Step-by-step explanation:

{x}^{2}  - 2x - 15

{x}^{2}  - 5x + 3x - 15 = 0

x(x - 5) + 3(x - 5) = 0

(x - 5)(x + 3) = 0

x = -3 (Not possible)

x = 5

So, x = 5

8 0
2 years ago
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