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Phantasy [73]
3 years ago
10

The formula for finding the length of an arc on a circle is L=2πr(x360) , where r is the radius of the circle and x is the measu

re of the central angle of the arc. Solve for r. r=xL720π r=720πxL r=πx180L r=180Lπx
Mathematics
2 answers:
horrorfan [7]3 years ago
6 0

The formula for length of an arc on a circle is given by the formula:

L = \frac{2 \pi rx}{360},

where 'r' is the radius of the circle and 'x' is the measure of the central angle of the arc.

We have to determine the value of radius 'r'.

Since, L = \frac{2 \pi rx}{360}

By Cross multiplication, we get

360 \times L = 2 \pi rx

\frac{360 \times L}{ 2 \pi x} =r

\frac{180 \times L}{ \pi x} =r

r = \frac{180L}{ \pi x}

Therefore, the radius 'r' is given by r = \frac{180L}{ \pi x}.

Option 4 is the correct answer.

Dmitry [639]3 years ago
4 0

I got the opposite of this answer, it was instead "solve for x."

With anyone with this for their question, the answer is the opposite:

x = 180L/\pir

Thus x = 180L/pi<em> r</em>, looks just like D, but the "r" and "x" placements are switched.

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denpristay [2]
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In Jacks class, 75% of the students are girls. there are 18 girls in the class. what is the total number if students in the clas
Andre45 [30]
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/.75   /.75 (divide both sides by .75 to isolate x)

24 = x

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6 0
3 years ago
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the simple interest paid on a loan over a period of two and a half years is birr 375. if the rate of interest was 8% per annum,
Reil [10]

Answer:

4678.5

Step-by-step explanation:

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8 0
3 years ago
What is the greatest integer a, such that a^2 + 3b is less than (2b)^2, assuming that b is 5? Please answer and have a nice day!
solniwko [45]
<h3>Answer: Largest value is a = 9</h3>

===================================================

Work Shown:

b = 5

(2b)^2 = (2*5)^2 = 100

So we want the expression a^2+3b to be less than (2b)^2 = 100

We need to solve a^2 + 3b < 100 which turns into

a^2 + 3b < 100

a^2 + 3(5) < 100

a^2 + 15 < 100

after substituting in b = 5.

------------------

Let's isolate 'a'

a^2 + 15 < 100

a^2 < 100-15

a^2 < 85

a < sqrt(85)

a < 9.2195

'a' is an integer, so we round down to the nearest whole number to get a \le 9

So the greatest integer possible for 'a' is a = 9.

------------------

Check:

plug in a = 9 and b = 5

a^2 + 3b < 100

9^2 + 3(5) < 100

81 + 15 < 100

96 < 100 .... true statement

now try a = 10 and b = 5

a^2 + 3b < 100

10^2 + 3(5) < 100

100 + 15 < 100 ... you can probably already see the issue

115 < 100 ... this is false, so a = 10 doesn't work

5 0
3 years ago
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