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svet-max [94.6K]
2 years ago
7

Do you guys like my pfp? (and do you know what it is?)

Mathematics
1 answer:
Sladkaya [172]2 years ago
4 0

Yes, I do like your pfp it looks very cool

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Guess the number pleaseeee​
MaRussiya [10]
I might be wrong but 9

9*2=18
18+7= 25
25 square root = 5

Answer is 9
7 0
3 years ago
CAN SOMONE PLS HEP WITH THIS!!!! ASAP FOR BRAINLIEST
quester [9]

Answer: How can I help?

Step-by-step explanation:

6 0
3 years ago
sume of angles of a triangel greater than 180 degrees when teh triange is so large that it extends to cosmological scales
Vesna [10]

In hyperbolic geometry, the angle sum of a triangle is always less than 180 degrees.

A lune is a wedge of a sphere with angle θ, represented by L(θ) in the proof.

α, β, and γ are the three angles of the triangle.

4πr²+4area[αβγ]=2L(α)+2L(β)+2L(γ)

2(2πr²+2area[αβγ])=2(L(α)+L(β)+L(γ))

2πr²+2area[αβγ]=L(α)+L(β)+L(γ)

At this point, we need to use a theorem that states that a lune whose corner angle is θ radians has an area of 2θr².

2πr²+2area[αβγ]=2αr²+2βr²+2γr²

2πr²+2area[αβγ]=2r²(α+β+γ)

π+area[αβγ]r²=α+β+γ

At this point, it is clear that the sum of the angles is equal to π plus the area[αβγ]r² (which cannot be zero).

To learn more about triangles and geometry,

brainly.com/question/2773823

#SPJ4

5 0
11 months ago
Matt is running in a 9 mile race over
iren2701 [21]

Answer:54 minutes

Step-by-step explanation:

9 x 5 = 45 then 45+ the.1 minute from each mile so it would be 9 +45 = 54 minutes

4 0
3 years ago
  What values for θ (0 ≤ θ ≤ 2pie ) satisfy the equation? 
svetoff [14.1K]
Factor out the cos<span>θ:
</span>cosθ (2sin<span>θ + sqrt3) = 0
</span>Therefore, the only ways this can happen are if either cosθ = 0 or if (2sin<span>θ + sqrt3) = 0
</span>The first case, cosθ = 0 only at θ <span>= pi/2, 3pi/2.
</span>The second case, <span>(2sin<span>θ + sqrt3) = 0 simplifies to:
</span></span>sin<span>θ = (-sqrt3)/2
</span><span><span>θ = 4pi/3, 5pi/3
</span></span><span><span>Therefore the answer is A.
</span></span>
7 0
3 years ago
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