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DerKrebs [107]
2 years ago
14

There is a a circle whose circumference is 34.54 cm.

Mathematics
1 answer:
Naily [24]2 years ago
5 0

Answer:

\huge\boxed{Diameter=11cm}

Step-by-step explanation:

Important information:

Circumference of a circle= \pi*diameter

Method:

To work out the diameter, we would need to divide the circumference of 34.54 by pi, which in this case is 3.14, this gives us 11cm. This is because we are simply rearanging the formula of the circumference of a circle to make the diameter the subject.

Step By Step:

1) Rearange the formula to make the diameter the subject.

diameter=\frac{circumference}{\pi}

2) Divide 34.54 by 3.14(\pi).

34.54/3.14=11cm

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Given the arithmetic sequence with the following two
kondor19780726 [428]

Answer:

aₙ = -15 + 10 * (n-1)

Step-by-step explanation:

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-x + 2y = 3<br> 2x – 3y = -6
s2008m [1.1K]

Answer:

x = -3

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Step-by-step explanation:

<u>Given</u><u> </u><u>equations</u><u> </u><u>:</u><u>-</u><u> </u>

<u>-x</u><u> </u><u>+</u><u> </u><u>2</u><u>y</u><u> </u><u>=</u><u> </u><u>3</u><u> </u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>i</u><u> </u><u>)</u>

<u>2</u><u>x</u><u> </u><u>-</u><u> </u><u>3</u><u>y</u><u> </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>ii</u><u> </u><u>)</u>

<u>From</u><u> </u><u>(</u><u> </u><u>i</u><u> </u><u>)</u><u> </u><u> </u>

<u>-x</u><u> </u><u>+</u><u> </u><u>2</u><u>y</u><u> </u><u>=</u><u> </u><u>3</u><u> </u>

<u>-x</u><u> </u><u>=</u><u> </u><u>3</u><u> </u><u>-</u><u> </u><u>2</u><u>y</u><u> </u>

<u>x</u><u> </u><u>=</u><u> </u><u>2</u><u>y</u><u> </u><u>-</u><u> </u><u>3</u><u> </u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>iii</u><u> </u><u>)</u>

<u>From</u><u> </u><u>(</u><u> </u><u>ii</u><u> </u><u>)</u><u> </u>

<u>2</u><u>x</u><u> </u><u>-</u><u> </u><u>3</u><u>y</u><u> </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u>

<u>2</u><u>x</u><u> </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u><u>+</u><u> </u><u>3</u><u>y</u><u> </u>

<u>x =  \frac{ - 6 + 3y}{2}</u>

<u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>iv</u><u> </u><u>)</u>

<u>Equating</u><u> </u><u>(</u><u> </u><u>iii</u><u> </u><u>)</u><u> </u><u>and</u><u> </u><u>(</u><u> </u><u>iv</u><u> </u><u>)</u>

<u>x</u><u> </u><u>=</u><u> </u><u>x</u><u> </u>

<u>2y - 3 =  \frac{ - 6 + 3y}{2}</u>

4y - 6 = -6 + 3y

4y - 3y = -6 + 6

y = 0

Putting value of y in ( iii )

x = 2y - 3

x = 2 ( 0 ) - 3

x = -3

4 0
2 years ago
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