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1 year ago

HELP ASAP!!!The circumference of a circle is 6 pi meters. Find the radius of the circle

Mathematics

1 answer:

tatiyna1 year ago

6 0

Step-by-step explanation:

radius is just the diameter cut in half.

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I need help solving this!

blagie [28]

Answer:

miles hybrid car went = 9.80 gal × 54.1 miles/gal

= 530.18 miles

132km × 0.621 = 81.972 miles

530.18 miles = 9.80 gal

1 mile = 9.80/530.18

= 0.018 gal

81.972 miles = 0.018 × 81.972

= 1.515 gal

1.515 gal × 3.785 = 5.734 litres

3 0

3 years ago

√144² = <br>√9 + √16 = <br>√9 + √16 × √400 =

Morgarella [4.7K]

Answer:

Step-by-step explanation

√144²

12²

144

√9 + √16

3 + 4

√9 + √16 × √400 =

3 + 2^{4} x 5

3 + 80

7 0

3 years ago

The scale drawing has a scale of 1/2 in: 8 mi. Find the length on the drawing for 2 in Please answer asap

11111nata11111 [884]

Answer:

32 mi

Step-by-step explanation:

Solve using proportions.

$\frac{\frac{1}{2}in}{8 mi} =\frac{2in}{y}$

Find the scale factor (how to get from left to right)

To get from left numerator to right numerator, multiply by 4.

(1/2) X 4 = 2

The scale factor is 4.

Multiply the left denominator by the scale factor to get "y".

8 mi X 4 = 32 mi

Therefore 2 inches represent 32 miles.

3 0

3 years ago

Pls help me with this <br>ans should be well explained

Snezhnost [94]

Let's observe

#1st column

3+2+1+4=10

10th alphabet=J

#2nd

5+9+4+6=24

24th alphabet=X

#3rd

2+1+3+3=9

9 th Alpha bet=I

#4th

8+4+9+5=26

26th Alphabet=Z

#5th

7+3+1+2=13

13 th alphabet=M

Option B is correct

5 0

3 years ago

A consumer organization estimates that over a 1-year period 20% of cars will need to be repaired once, 8% will need repairs

Sladkaya [172]

Answer:

Probability that a car need to be repaired once = 20% = 0.20

Probability that a car need to be repaired twice = 8% = 0.08

Probability that a car need to be repaired three or more = 2% = 0.02

a) If you own two cars what is the probability that neither will need repair?

Probability that a car need to be repaired once , twice and thrice or more= 0.20+0.08+0.02=0.3

Probability that car need no repair = 1-0.3=0.7

Neither car will need repair= $0.7 \times 0.7=0.49$

b) both will need repair?

Probability both will need repair = $0.3 \times 0.3=0.09$

c)at least one car will need repair

Neither car will need repair= $0.7 \times 0.7=0.49$

Probability that at least one car will need repair= 1-0.49 = 0.51

6 0

2 years ago