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seraphim [82]
3 years ago
6

Please help me, will mark you as brainiest! Don’t answer if you don’t know it!

Mathematics
1 answer:
Vesnalui [34]3 years ago
6 0

Answer:

The correct option is;

26.8

Step-by-step explanation:

The given parameters are;

The perpendicular bisectors of the sides of ΔMNP = Segments OQ, OR, and OS

Therefore;

in Segment MN, segment QM = segment QN, by definition of bisected line

Segment OQ = Segment OQ by reflexive property

∠OQM = ∠OQN = 90° by definition of perpendicular bisector

∴ ΔOQN ≅ ΔOQM, by SAS rule of congruency

Segment MO ≅ Segment ON by Congruent Parts of Congruent Triangle are Congruent, CPCTC

Segment MO = Segment ON = 26.8, by definition of congruency

Segment MO = 26.8

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Answer:

A. at least 5 sales

Step-by-step explanation:

if he is already making $20 dollars each day and he wants $50 on friday I first took 50-20 to find what he needs to make just in sales once I found out he needs $30 so then I divided 30 by 6 to find out exactly how many sales he needs to make.

7 0
3 years ago
Please Help!!
gladu [14]

Annual rate: 500*0.07 = 35
Year 1: 35
Year 2: 35
Year 3: 35
Year 4: 35
Year 5: 35+500=535
Total=675

Compounded annually
Year 1: 500*1.05=525
Year 2: 525*1.05=551.25
Year 3: 551.25*1.05=578.8125=578.81
Year 4: 578.8125*1.05=607.753125=607.75
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4 years ago
A. -22<br> B. 11<br> C. 13<br> D. 26
AlexFokin [52]

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2n - 3  : 8

5 0
3 years ago
Several different cheeses are for sale. The cheese comes in wedges
Tpy6a [65]

Answer:

39.30 cm^{2}

Step-by-step explanation:

The area of the top surface of the wedge is the same as the area of a sector.

Area of a sector = (θ/360^{o})\pir^{2}

where: θ is the central angle, and r is the radius.

Given that: central angle = 45^{o}, radius = 10 centimetre.

Area of the top surface = Area of a sector = (θ/360^{o})\pir^{2}

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                                      = \frac{1}{8} x \frac{22}{7} x 100

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The area of the top surface of the wedge is 39.30 cm^{2}.

3 0
3 years ago
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son4ous [18]

Answer:

11

Step-by-step explanation:

8 0
3 years ago
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