Which graph best represents the equation 3x + 4y = 12?
1 answer:
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Answer:
Part A:
m∠VHT = 152°
Part B:
m∠QTS = 152°
Part C:
m∠ZHQ = 28°.
Step-by-step explanation:
Part A:
The given parameters are;
m∠HXU = 113°
Segment BQ and segment UD intersect at m∠XAT = 95°
We have that m∠HXU + m∠HXS = 180° (Angles on a straight line)
Therefore;
m∠HXU = 180° - m∠HXS = 180° - 113° = 67°
m∠HXU = 67°
m∠XAT + m∠XAH = 180° (Angles on a straight line)
m∠XAH = 180° - m∠XAT = 180° - 95° = 85°
m∠XAH = 85°
In triangle XAH, we have;
m∠XAH + m∠HXU + m∠XHA = 180° (Angle sum property of a triangle)
∴ m∠XHA = 180° - (m∠XAH + m∠HXU) = 180° - (85° + 67°) = 28°
m∠XHA = 28°
m∠VHT + m∠XHA = 180° (Angles on a straight line)
m∠VHT = 180° - m∠XHA = 180° - 28° = 152°
m∠VHT = 152°
Part B:
m∠QTS ≅ m∠VHT (Corresponding angles are congruent)
∴ m∠QTS = 152° (Substitution property)
Part C:
m∠ZHQ ≅ m∠XHA (Reflexive property)
∴ m∠ZHQ = 28°.
Answer:
Step-by-step explanation:
The area of a regular octahedron is given by:
area = . Let a is the length of the edge (diagonal).
area =
Given that the diagonal of the octahedron is equal to height (h) of the tetrahedron i.e.
a = h, where h is the height of the tetrahedron and a is the diagonal of the octahedron. Let the edge of the tetrrahedron be e. To find the edge of the tetrahedron, we use:
The area of a tetrahedron is given by:
area = =
The ratio of area of regular octahedron to area tetrahedron regular is given as:
Ratio =