The minute hand of a clock is 6 inches long. How far does the tip of the minute hand move in 35 minutes?

Answers

Answer 1
Answer:

Answer:

22 in

Step-by-step explanation:

It travels 35/60 ths of the complete circumference of a circle with r =6 in

diameter = 12     circumference = pi * d

35/ 60  * pi * 12 =  ~22 inches


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An air conditioning system can circulate 320 cubic feet of air per minute. How many cubic yards of air can it circulate per minute?

Answers

Step-by-step explanation:

It is given that, an air conditioning system can circulate 320 cubic feet of air per minute. We need to find how many cubic yards of air can it circulate per minute?

1 yard = 3 foot

1 feet = (1/3) yard

320 feet = 106.66 yard

320\ \text{feet}^3=320*( (1)/(3))^3\ \text{yard}^3\n\n=11.85\ \text{yard}^3

So, it will circulate 11.85 cubic yard of air per minute.

Final answer:

An air conditioning system can circulate 11.85 cubic yards of air per minute.

Explanation:

To convert cubic feet to cubic yards, we need to divide the given value by 27 since 1 cubic yard is equal to 27 cubic feet. So, 320 cubic feet is equal to 320/27 = 11.85 cubic yards. Therefore, the air conditioning system can circulate 11.85 cubic yards of air per minute.

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Simplify 5/6i please help

Answers

Answer:

5/6 can´t be simplified

Step-by-step explanation:

Answer:

5i/6

Step-by-step explanation:

(5)/(6)i\n\n\mathrm{Multiply\:fractions}:\quad \:a\cdot (b)/(c)=(a\:\cdot \:b)/(c)\n=(5i)/(6)

Which expression is equivalent to 2la+26)-2-26?

Answers

Answer:

Dear user,

Answer to your query is provided below

2(al−1) is the expression equivalent for 2la+26−2−26

Convert to fractions and write in simplest form.A) 0.02
B) 12%
C) 0.5%
D) 1.12

Answers

Answer:

Given below.

Explanation:

A) 0.02 = (2)/(100) =  (1)/(50)

B) 12% = (12)/(100) = (6)/(50) = (3)/(25)

C) 0.5% = (0.5)/(100) = (1)/(200)

D) 1.12 = (28)/(25)

If a new truck costs $43,750 and it depreciates 18% per year, what will the truck be worth in 5 years? 1 SEE ANSWER

Answers

if it depreciates by 18%, then as each year passes, its worth is reduced to 82% of its value
So, after 5 years, it is worth 43750

B(n)=2^n A binary code word of length n is a string of 0's and 1's with n digits. For example, 1001 is a binary code word of length 4. The number of binary code words, B(n), of length n, is shown above. If the length is increased from n to n+1, how many more binary code words will there be? The answer is 2^n, but I don't get how they got that answer. I would think 2^n+1 minus 2^n would be 2. Please help me! Thank you!

Answers

Answer:

More number of words that can be made: \bold{2^n}

Please refer to below proof.

Step-by-step explanation:

Given that:

The number of binary code words that can be made:

B(n)  =2^n

where n is the length of binary numbers.

Binary numbers means 2 possibilities either 0 or 1.

Here, suppose if we have 5 as the length of binary number.

And there are 2 possibilities for each digit.

So, total number of possibilities will be 2* 2* 2* 2* 2 = 2^5

If the length of binary number is 2.

The total words possible are 2^2.

These numbers are:

{00, 01, 10, 11}

If the length of binary number is 3. (increasing the 'n' by 1)

The total words possible are 2^3.

These words are:

{000, 001, 010, 100, 011, 101, 110, 111}

So, number of More binary words = 8 - 4 = 4 or 2^2 or 2^n.

So, the answer is 2^n.

Let us try to prove in generic terms:

B(n) = 2^n

Increasing the n by 1:

B(n+1) = 2^(n+1)

Number of more words made by increasing n by 1:

B(n+1) -B(n)= 2^(n+1) -2^n\n\Rightarrow 2* 2^(n) -2^n\n\Rightarrow 2^n(2-1)\n\Rightarrow \bold{2^n}

Hence, proved that:

More number of words that can be made: \bold{2^n}

Final answer:

When the length of a binary code word increases from n to n+1, the number of additional binary code words is equal to the number of binary code words of length n, which is 2^n.

Explanation:

When the length is increased from n to n+1, the number of binary code words of length n+1 is equal to the number of binary code words of length n multiplied by 2. This is because for each binary code word of length n, we can append a 0 or a 1 to create two new binary code words of length n+1. Therefore, the number of additional binary code words is equal to the number of binary code words of length n, which is2^n.

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