The figure is missing so I attached a helping figure
Answer:
Line segment ST is congruent to line segment UT
Step-by-step explanation:
From the attached figure
∵ ST and UT are tangents to circle K at points S and U
∵ SK and UK are radii in the circle K
- The tangent is perpendicular to the radius at the point of tangent
∴ ST ⊥ KS ⇒ at point S
∴ m∠KST = 90°
∴ UT ⊥ KU ⇒ at point U
∴ m∠KUT = 90°
∴ m∠KST = m∠KUT
In the two triangles KST and KUT
∵ KS = KU ⇒ radii
∵ m∠KST = m∠KUT ⇒ proved
∵ KT is a common side in the two triangles
- That means the two triangles are congruent by HL postulate
of congruence (hypotenuse and leg of right triangle)
∴ Δ KST ≅ KUT ⇒ HL postulate of congruence
- By using the result of congruence
∴ ST = UT
Line segment ST is congruent to line segment UT
Step-by-step explanation:
The mean absolute deviation is:
MAD = ∑ᵢ₌₁ⁿ│xᵢ − μ│/ n
First, find the mean.
μ = (x + 2x + 3x + 4x) / 4
μ = 2.5x
Now find the mean absolute deviation:
MAD = (│x − 2.5x│+│2x − 2.5x│+│3x − 2.5x│+│4x − 2.5x│) / 4
MAD = (│-1.5x│+│-0.5x│+│0.5x│+│1.5x│) / 4
MAD = (1.5x + 0.5x + 0.5x + 1.5x) / 4
MAD = (4x) / 4
MAD = x
Answer:
the answer is 84
Step-by-step explanation:
(4(1) +10)+8(5)-10(-3)
4+10+40+30
=84
Answer:
7
Step-by-step explanation:
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