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Sunny_sXe [5.5K]
4 years ago
9

Which set of three numbers could be the side lengths of a triangle?

Mathematics
1 answer:
Tresset [83]4 years ago
5 0

Answer:

<h2>А. 3, 5, 7</h2><h2 />

Step-by-step explanation:

Because

7 - 3 = 4

7 + 3 = 10

then

7 - 3 < 5 < 7 + 3

the 3 other cases don’t fit with this rule.

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rewona [7]

Answer:

Plug 3 into everywhere there is an 'x', and solve.

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You have 6 bracelets and 15 necklaces. Find the number of ways you can wear one bracelet and one necklace
KonstantinChe [14]
The answer would be 90
5 0
4 years ago
Can u help please :)
Lyrx [107]

Answer:

z = 8

Step-by-step explanation:

All angles inside a triangle add up to 180 so:

90 + 40 + 5z + 10 = 180

5z + 140 = 180

5z = 40

z = 8

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3 years ago
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vitfil [10]
2450 is the correct answer because if you multiply 3500 x .70 then you get 2450. Hope this helped 
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3 years ago
Read 2 more answers
42:28
gogolik [260]

Answer:

The statements about arcs and angles that are true in the figure are;

1) ∠EFD ≅ ∠EGD

2) \overline{ED}\cong \overline{FD}

3) mFD = 120°

Step-by-step explanation:

1) ∠ECD + ∠CEG + ∠CDG + ∠GDE = 360° (Sum of interior angle of a quadrilateral)

∠CEG = ∠CDG = 90° (Given)

∠GDE = 60° (Given)

∴ ∠ECD = 360° - (∠CEG + ∠CDG + ∠GDE)

∠ECD = 360° - (90° + 90° + 60°) = 120°

∠ECD = 2 × ∠EFD (Angle subtended is twice the angle subtended at the circumference)

120° = 2 × ∠EFD

∠EFD = 120°/2 = 60°

∠EFD ≅ ∠EGD

∠ECD = 120°

∠EGD = 60°

∴∠EGD ≠ ∠ECD

2) Given that arc mEF ≅ arc mFD

Therefore, ΔECF and ΔDCF are isosceles triangles having two sides (radii EC and CF in ΔECF and radii EF and CD in ΔDCF

∠ECF = mEF = mFD = ∠DCF (Given)

∴ ΔECF ≅ ΔDCF (Side Angle Side, SAS, rule of congruency)

\\ \overline{EF}\cong \overline{FD} (Corresponding Parts of Congruent Triangles are Congruent, CPCTC)

∠FED ≅ ∠FDE (base angles of isosceles triangle)

∠FED + ∠FDE + ∠EFD = 180° (sum of interior angles of a triangle)

∠FED + ∠FDE = 180° - ∠EFD = 180° - 60° = 120°

∠FED + ∠FDE = 120° = ∠FED + ∠FED (substitution)

2 × ∠FED  = 120°

∠FED = 120°/2 = 60° = ∠FDE

∴ ∠FED = ∠FDE = ∠EFD =  60°

ΔEFD  is an equilateral triangle as all interior angles are equal

\\ \overline{EF}\cong \overline{FD}\cong \overline{ED} (definition of equilateral triangle)

\overline{ED}\cong \overline{FD}

3) Having that ∠EFD = 60° and ∠CFE = ∠CFD (CPCTC)

Where, ∠EFD = ∠CFE + ∠CFD (Angle addition)

60° = ∠CFE + ∠CFD = ∠CFE + ∠CFE (substitution)

60° = 2 × ∠CFE

∠CFE =60°/2 = 30° = ∠CFD

\overline{CF}\cong \overline{CD} (radii of the same circle)

ΔFCD is an isosceles triangle (definition)

∠CFD ≅ ∠CDF (base angles of isosceles ΔFCD)

∠CFD + ∠CDF + ∠DCF = 180°

∠DCF = 180° - (∠CFD + ∠CDF) = 180° - (30° + 30°) = 120°

mFD = ∠DCF (definition)

mFD = 120°.

5 0
3 years ago
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