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2 years ago

The point (1, 4) lies on a circle that is centered at (1, 1). Which statements are correct? Check all that apply.

Mathematics

2 answers:

Semenov [28]2 years ago

6 0

the circle's radius is 3 units

the point (- 2, 1 ) lies on the circle

the equation of a circle in standard form is

(x - a)² + (y - b)² = r²

where (a, b) are the coordinates of the centre and r is the radius

the radius is the distance from the centre to the point (1, 4 ) on the circle

using (1, 4) and (1,1) in the distance formula, then

r = √(1 - 1 )² + (1 - 4)² = √(0 + 9) =√9 = 3 ⇒ r² = 9

(x - 1 )² + (y - 1 )² = 9 ← equation of circle

substitute the given points into the equation and if equation is true then they lie on the circle

(- 2, 1 ) : (- 2 - 1 )² + (1 - 1 )² = 9 + 0 = 9 ← true

Hence (- 2, 1 ) lies on the circle

(3, 3 ) : (3 - 1 )² + (3 - 1 )² = 4 + 4 = 8 ≠ 9

(3, 3 ) does not lie on the circle

Monica [59]2 years ago

4 0

Answer: b and d

Step-by-step explanation:

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klemol [59]

Answer:

7.5

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Two quadrilaterals are congruent. One has vertices P, N, O, and M, and the other has vertices S, T, V, and U. These correspondin

Bogdan [553]

Answer: $NPOM \cong VUTS$ and $OPNM \cong TVUS$ will be correct.

Explanation:

Given: two quadrilaterals having verticals P, N, O,M and S,T,V,U are congruent, where, OM is congruent or equal to TS and $\angle P\cong \angle U$ .

in quadrilaterals NPOM and VUTS-

since, the condition $\angle P = \angle U$

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then we can say according to the property of quadrilateral, their corresponding sides must be congruent. so they are congruent.

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we can understand it by making the figures.

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1 year ago

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Alexeev081 [22]

so the answer is 36.

Hope this will help u...

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David proved the Law of Cosines with reference to ∆ABC using a different method. Arrange the steps of his proof in the correct s

AURORKA [14]

Answer: We can arrange the steps with help of below explanation.

Step-by-step explanation:

Here ABC is a triangle,

Draw a perpendicular from BD to side AC ( construction)

where $D\in AC$

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cos C =CD/ BC

CD= a cos C

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DA= b-acos C

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BD = a sin C

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$c^2 =BD^2 +DA^2$

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⇒ $c^2 =a^2sin ^2C+b^2-2ab cosC+a^2 cos ^2C$ ( On simplification)

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⇒ $c^2=a^2(sin^2C+cos^2C) +b^2-2abcosC$

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2 years ago