1 1 comp supl word prob
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PROBLEM 1PROBLEM 3
PROBLEM 2PROBLEM 4
PROBLEM 5
PROBLEM 8PROBLEM 7PROBLEM 6
STANDARD 13
SUPPLEMENT AND COMPLEMENT: NUMERIC
PROBLEM 10PROBLEM 9
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SUPPLEMENT: GENERAL
COMPLEMENT: GENERAL
SUPPLEMENT AND COMPLEMENT: GENERAL
WORD PROBLEMS:
SUPPLEMENTARY AND COMPLEMENTARY ANGLES
END SHOW
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STANDARD 7:
Students prove and use theorems involving the properties of parallellines cut by a transversal, the properties of quadrilaterals, andproperties of circles.
ESTÁNDAR 7:
Los estudiantes prueban y usan teoremas involucrando las propiedadesde líneas paralelas cortadas por una transversal, las propiedades de
cuadriláteros, y las propiedades de círculos.
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STANDARD 13
Find the measure of the COMPLEMENT and SUPPLEMENT of the angle below:
90° ± 65° = 25°65°
115°
Supplement:
180° ± 65° = 115°
Calculating the complement:
Calculating the supplement:
25°
Complement
65°
65°
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Find the measure of the COMPLEMENT and SUPPLEMENT of the angle below:
90° ± 55° = 35°55°
125°
Supplement:
180° ± 55° = 125°
Calculating the complement:
Calculating the supplement:
STANDARD 13
35°
Complement
55°
55°
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X°
Find the measure of the SUPPLEMENT of the angle below:
180° ± X°
(180 ± X)°
Supplement:
STANDARD 13
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Find the measure of the COMPLEMENT of the angle below:
90 ± X°
Complement
(90-X)°
X°
STANDARD 13
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Find COMPLEMENT AND SUPPLEMENT of any given angle:
90°± X°
(180 ± X)°
Supplement:
180° ± X°
Calculating the complement:
Calculating the supplement:
(90 ± X)°
Complement
X°
X°
Let¶s call our angle X:
X°
STANDARD 13
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X is the angle
180 ± X is the supplement
180 ± X =X ± 20
The supplement of an angle is twenty less than the angle, find this angleand the supplement.
+ X + X
180 = 2X ± 20+20 +20
200 = 2X
2 2X = 100 Then the supplement:
180 ± X = 180 ± 100
= 80
The angle is 100° and the supplement is 80°.
X180 ± X
STANDARD 13
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X is the angle
90 ± X is the complement
90 ± X = 2X
The complement of an angle is two times the angle, find this angle and thecomplement.
+ X + X
3 3
X = 30 Then the complement:
90 ± X = 90 ± 30= 60
The angle is 30° and the complement is 60°.
90 = 3X
X
90 - X
STANDARD 13
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X is the angle
90 ± X is the complement
90 ± X =X + 10
The complement of an angle is 10 more than the angle, find this angle andthe complement.
+X +X
90 = 2X + 10-10 -10
80 = 2X
2 2X =40 Then the complement:
90 ± X = 90 ± 40= 50
The angle is 40° and the complement is 50°.
X
90 - X
STANDARD 13
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X is the angle
180 ± X is the supplement
180 ± X =
The supplement of an angle is a third of the angle, find this angle and thesupplement.
540 = 4X4 4X = 135
Then the supplement:
180 ± X = 180 ± 135
= 45
The angle is 135° and the supplement is 45°.
X180 ± X
X13
180 ± X = X13
33
540 ± 3X = X
+3X +3X
STANDARD 13
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X is the angle
180 ± X is the supplement
180 ± X =
The supplement of an angle is a fourth of the angle, find this angle and thesupplement.
720 = 5X5 5X = 144
Then the supplement:
180 ± X = 180 ± 144
= 36
The angle is 144° and the supplement is 36°.
X180 ± X
X14
180 ± X = X14
44
720 ± 4X = X
+4X +4X
STANDARD 13
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X is the angle
90 ± X is the complement
90 ± X = 3X ± 10
The complement of an angle is 10 less than three times the angle, find thisangle and the complement.
+X +X
90 = 4X ± 10+10 +10
100 = 4X
4 4X = 25 Then the complement:
90 ± X = 90 ± 25= 65
The angle is 25° and the complement is 65°.
X
90 - X
STANDARD 13
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X is the angle
180 ± X is the supplement
180 ± X = 3X + 40
The supplement of an angle is forty more than the triple of the angle, findthis angle and the supplement.
+ X + X
180 = 4X + 40-40 -40
140 = 4X
4 4X = 35 Then the supplement:
180 ± X = 180 ± 35
= 145
The angle is 35° and the supplement is 145°.
X180 ± X
STANDARD 13
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X is the angle
90 ± X is the complement
90 ± X = 2X ± 30
The complement of an angle is thirty less than twice the angle, find thisangle and the complement.
+X +X
90 = 3X ± 30+30 +30
120 = 3X
3 3X = 40 Then the complement:
90 ± X = 90 ± 40= 50
The angle is 40° and the complement is 50°.
X
90 - X
STANDARD 13
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X is the angle
180 ± X is the supplement
5(180 ± X) = (90 ± X) + 630
Five times the supplement of an angle is 630 more than its complement.Find the angle, the supplement and the complement.
+5X +5X
900 = 4X + 720-720 -720
180 = 4X4 4
X = 45
Then the supplement:
180 ± X = 180 ± 45
= 135
The angle is 45°, the supplement is 135° and the complement is 45°.
900 ± 5X = 720 ± X
Then the complement:
90 ± X = 90 ± 45
= 45
STANDARD 13
90 ± X is the complementX180 ± X
X
90 - X
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