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alekssr [168]
3 years ago
6

What is the area of a circle with circumference of 12π?

Mathematics
2 answers:
ad-work [718]3 years ago
6 0

Answer:A≈11.46


Step-by-step explanation:


Ludmilka [50]3 years ago
5 0
The answer is 11.46
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Solve the system by the elimination method.
Vika [28.1K]
10x + 2y - 6 = 0

Multiply 5x + y - 3 = 0 by 2

2(5x + y - 3 = 0) = 10x + 2y - 6 = 0
7 0
3 years ago
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What is the ph of 0.1M HNO3
11111nata11111 [884]

Answer:

i hope this helps

Step-by-step explanation:

the pH of the solution is 5.74.

5 0
4 years ago
Find 3 consecutive odd integers whose sum is 225, explain please
aleksklad [387]
3 odd integers have a property that they average up to the middle large one. Let's say we have 3, 5, and 7. 3 is 2 less than 5 and 7 is 2 more than 5. so when you add them it equals 2 times 5. 
After we know that, the sum of 3 odd integers is just 3 times the middle number. ex. 3+5+7 = 3 times 5 = 15
Then we know the some number times three = 225. we find out that the middle number is 75, so the other two are 73 and 77
5 0
4 years ago
How many terms are there in the sequence 1, 8, 28, 56, ..., 1 ?
BabaBlast [244]

Answer:

9 terms

Step-by-step explanation:

Given:  

1, 8, 28, 56, ..., 1

Required

Determine the number of sequence

To determine the number of sequence, we need to understand how the sequence are generated

The sequence are generated using

\left[\begin{array}{c}n&&r\end{array}\right] = \frac{n!}{(n-r)!r!}

Where n = 8 and r = 0,1....8

When r = 0

\left[\begin{array}{c}8&&0\end{array}\right] = \frac{8!}{(8-0)!0!} = \frac{8!}{8!0!} = 1

When r = 1

\left[\begin{array}{c}8&&1\end{array}\right] = \frac{8!}{(8-1)!1!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8

When r = 2

\left[\begin{array}{c}8&&2\end{array}\right] = \frac{8!}{(8-2)!2!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} =2 8

When r = 3

\left[\begin{array}{c}8&&3\end{array}\right] = \frac{8!}{(8-3)!3!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56

When r = 4

\left[\begin{array}{c}8&&4\end{array}\right] = \frac{8!}{(8-4)!4!} = \frac{8!}{4!3!} = \frac{8 * 7 * 6 * 5 * 4!}{4! *4*3* 2 *1} = \frac{8 * 7 * 6*5}{4*3 *2 *1} = 70

When r = 5

\left[\begin{array}{c}8&&5\end{array}\right] = \frac{8!}{(8-5)!5!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56

When r = 6

\left[\begin{array}{c}8&&6\end{array}\right] = \frac{8!}{(8-6)!6!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} = 28

When r = 7

\left[\begin{array}{c}8&&7\end{array}\right] = \frac{8!}{(8-7)!7!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8

When r = 8

\left[\begin{array}{c}8&&8\end{array}\right] = \frac{8!}{(8-8)!8!} = \frac{8!}{8!0!} = 1

The full sequence is: 1,8,28,56,70,56,28,8,1

And the number of terms is 9

3 0
3 years ago
Sheila has 11
RSB [31]

Answer:

6

Step-by-step explanation:

11 1/2 = 11.5

2 1/4 = 2.25

11.5 ÷ 2.25= 5.11

she needs 6 containers because you round up 5.11 to 6.

if your doing the work in fractions rather decimals let me know and I can explain it that way too

5 0
3 years ago
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