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Nady [450]
2 years ago
11

NAME THE PARTS OF THE CIRRCLR​

Mathematics
1 answer:
Alexeev081 [22]2 years ago
5 0
1) O
2) OC
3) AB
4) EB, CB, AC, DA, CE and more others
5) DE
6) AB it’s half of a circle
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What is 2\20 divided by 2
Valentin [98]
\frac{2}{20} \div 2 \\ \\  \frac{1}{10} \div 2 \\ \\  \frac{1}{10} \times  \frac{1}{2} \\ \\  \frac{1 \times 1}{10 \times 2} \\ \\  \frac{1}{20} \\ \\

The final result is: 1/20 or 0.05.
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Find the quotient 22,080 ÷ 24?
Nikolay [14]

Answer:920

Step-by-step explanation:

plug it in a calculator

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Please I need help on this ASAP
KATRIN_1 [288]

Answer:

\frac{1}{256}

Step-by-step explanation:

\frac{1}{16}  \times  \frac{1}{16} =  \frac{1}{256}

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3 years ago
A petrol kiosk p is 12 km due north of another petrol kiosk q. The bearing of a police station r from p is 135 degree and that f
tiny-mole [99]

Answer:

Distance between P and R is 40.15 km.

Step-by-step explanation:

From the picture attached,

Petrol kiosk P is 12 km due North of another petrol kiosk Q.

Bearing of a police station R is 135° from P and 120° from Q.

m∠QPR = 180° - 135° = 45°

m∠PQR = 120°

m∠PRQ = 180° - (m∠QPR +m∠PQR)

             = 180° - (45° + 120°)

             = 180° - 165°

             = 15°

Now we apply sine rule in ΔPQR to measure the distance between P and R.

\frac{\text{sin}(\angle QPR)}{\text{QR}}= \frac{\text{sin}(\angle PQR)}{\text{PR}}=\frac{\text{sin}\angle PRQ}{\text{PQ}}

\frac{\text{sin}(45)}{\text{QR}}= \frac{\text{sin}(120)}{\text{PR}}=\frac{\text{sin}(15)}{\text{12}}

\frac{\text{sin}(120)}{\text{PR}}=\frac{\text{sin}(15)}{\text{12}}

PR = \frac{12\text{sin}(120)}{\text{sin}(15)}

PR = 40.15 km

Therefore, distance between P and R is 40.15 km.

8 0
3 years ago
Simplified form of 11[(9^2-5^2)/2^2+8]<br> A.224.5<br> B.242<br> C.51.3<br> D.110
aleksandr82 [10.1K]
B. 242
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4 years ago
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