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NARA [144]
3 years ago
5

Will give brainiest for this

Mathematics
1 answer:
cluponka [151]3 years ago
3 0

Answer:

lol, u can still ask the question

Step-by-step explanation:

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Misha bought 9 new baseball trading cards to add to her collection. The next day
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She started with 81 baseball cards.
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Sandra's cell phone plan gives her unlimited minutes for $15.50 per month. She is charged $0.08 for each text message, t. Which
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Answer:

c=15.50+0.08t

Step-by-step explanation:

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Which of the following methods would be the easiest to use to solve x2 + 4x + 3 = 0?
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It would be easier to do A
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Can someone help me on this pls? It’s urgent, so ASAP (it’s geometry)
GarryVolchara [31]

<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
2cos^2x = 1 <br><br> Solve for 0-360 degrees
Tems11 [23]

Answer:

45,135,315,225

Step-by-step explanation:

2cos^2x = 1

Divide each side by 2

cos^2x = 1/2

Take the square root of each side

sqrt( cos^2 x) = ±sqrt (1/2)

cos x  =±sqrt (1/2)

Make into two separate equations

cos x  =sqrt (1/2)    cos x = - sqrt(1/2)

Take the inverse cos of each side

cos ^-1 cos (x) = cos ^-1 (sqrt (1/2))   cos ^-1 cos (x) = cos ^-1 (-sqrt (1/2))

x = cos ^-1 (sqrt (1/2))   x = cos ^-1 (-sqrt (1/2))

x = 45  +360 n              x = 135+ 360n

x = 315+360 n               x =225+360n

Between 0 and 360

45,135,315,225

5 0
3 years ago
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