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krok68 [10]
3 years ago
8

Put the steps for deriving the formula for the arc length of a circle in the correct order.

Mathematics
1 answer:
allochka39001 [22]3 years ago
3 0
Just look it up that’s what I do
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16 Find the distance between the two points of J(-8, 0) and K(1, 4). Round to the nearest tenth. * (6 Points) O 10.3 o 8.5 O 11.
guajiro [1.7K]

Answer:

9.8

Step-by-step explanation:

8 0
3 years ago
Which of the following is the statement of the triangle inequality theorem?
kupik [55]

we know that

<u>The triangle inequality theorem</u> states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side

so

Let

a,b,c------> the length sides of a triangle

The theorem states that three conditions must be met

<u>case 1)</u>

a+b > c

<u>case 2)</u>

a+c > b

<u>case3)</u>

c+b > a

therefore

<u>the answer is the option</u>

B. The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

5 0
3 years ago
Read 2 more answers
Which of the following statements are true? Select all that apply.
wariber [46]

Answer:

1. No. it equals 24

2. Yes. it equals 18

3. Yes. it equals 90

4. No. it equals 44

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Tan theta if sec theta= 5/2 and csc theta &lt; 0?
aliya0001 [1]
it's to 2.5 cuz u divide 5/2=2.5
8 0
3 years ago
If you are allowed to use numbers 1 – 20 and need to choose the passcode of an exact 4 digit code, how many possibilities are th
VMariaS [17]

Since in a pass code, the placement of the digits is important, therefore this means that to solve for the total number of possibilities we have to make use of the principle of Permutation. The formula for calculating the total number of possibilities using Permutation is given as:

P = n! / (n – r)!

where,

n = is the total amount of numbers to choose from = 20

r = is the total number of digits needed in the passcode = 4

 

Therefore solving for the total possibilities P:

P = 20! / (20 – 4)!

P = 20! / 16!

P = 116,280

 

<span>Hence there are a total of 116,280 possibilities of pass codes.</span>

4 0
3 years ago
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