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s344n2d4d5 [400]
3 years ago
12

Solve the system by substitution  y=-382 y = 6x + 3 

Mathematics
1 answer:
nikdorinn [45]3 years ago
6 0

okay so i assume you meant these two:

y=-382 and y=6x+3

Answer:

y=-382

x=-385/6

Step By Step:

y=−382;y=6x+3

Step: Solve y=−382 for y:

y=−382

Step: Substitute −382 for y in y=6x+3:

y=6x+3

−382=6x+3

−382+−6x=6x+3+−6x (Add -6x to both sides)

−6x−382=3

−6x−382+382=3+382 (Add 382 to both sides)

−6x=385

Divide both sides by -6.

x=-385/6

y=-382

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Answer:

a) 240°

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Step-by-step explanation:

To solve these equations you have to use the inverse of the given trigonometric functions. The inverse of <em>sin</em> is <em>arcsin</em>, and the inverse of <em>tan </em>is <em>arctan. </em>Instead of giving an angle, what is its sine?, the question is: given a sine,  what is the angle?.

a)

sin(θ) = -√3/2

θ = arcsin(-√3/2)

θ = -60°

Given the periodicity of sine function, sin(-60°) is equivalent to sin(240°) (-60+180) and sin(300°) (-60+360).

b)

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θ = arctan(1/√3)

θ = 30°

c)

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csc(θ) = 1/sin(θ)

csc(θ) = -√2

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3 0
4 years ago
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Can someone help me on this pls? It’s urgent, so ASAP (it’s geometry)
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<u>Question 6</u>

1) \overline{AB} \cong \overline{BD}, \overline{CD} \perp \overline{BD}, O is the midpoint of \overline{BD}, \overline{AB} \cong \overline{CD} (given)

2) \angle ABO, \angle ODC are right angles (perpendicular lines form right angles)

3) \triangle ABO, \triangle CDO are right triangles (a triangle with a right angle is a right triangle)

4) \overline{BO} \cong \overline{OD} (a midpoint splits a segment into two congruent parts)

5) \triangle ABO \cong \triangle CDO (LL)

<u>Question 7</u>

1) \angle ADC, \angle BDC are right angles), \overline{AD} \cong \overline{BD}

2) \overline{CD} \cong \overline{CD} (reflexive property)

3) \triangle CDA, \triangle CDB are right triangles (a triangle with a right angle is a right triangle)

4) \triangle ADC \cong \triangle BDC (LL)

5) \overline{AC} \cong \overline{BC} (CPCTC)

<u>Question 8</u>

1) \overline{CD} \perp \overline{AB}, point D bisects \overline{AB} (given)

2) \angle CDA, \angle CDB are right angles (perpendicular lines form right angles)

3) \triangle CDA, \triangle CDB are right triangles (a triangle with a right angle is a right triangle)

4) \overline{AD} \cong \overline{DB} (definition of a bisector)

5) \overline{CD} \cong \overline{CD} (reflexive property)

6)  \triangle ADC \cong \triangle BDC (LL)

7) \angle ACD \cong \angle BCD (CPCTC)

8 0
2 years ago
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