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Varvara68 [4.7K]
3 years ago
14

Which of the factors of 120 are composite numbers?

Mathematics
1 answer:
Pani-rosa [81]3 years ago
6 0

Answer:

Find a common prime factor in each of these pairs of numbers

a.21 and 411

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Today is my exam so are you guys ready to help me?​
Vaselesa [24]

Answer:

we are always ready to help you

5 0
3 years ago
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A clock is positioned on an auditorium wall with its center 9 ft above the floor. The second hand on the clock is 10 inches long
oee [108]

Answer:

B : y=5/6cos(pi/30x)+9

Step-by-step explanation:

Edge 2020

7 0
4 years ago
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Find the first term of the arithmetic sequence in which a38= -5 and the common difference is -2.9
mestny [16]

Answer:

102.3

Step-by-step explanation:

a_38 = -5

difference d= -2.9

We use general formula

a_n = a_1 + (n-1)d

WE make the formula for a_38 th term

Plug in 38 for n

a_{38} = a_1 + (38-1)d

Now plug in -2.9 for d and -5 for a_38

-5 = a_1 + (38-1)(-2.9)

-5 = a_1 - 107.3

Now add 107.3 on both sides

102.3 = a_1

option A is correct



3 0
3 years ago
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Does anybody know how to do time with exponential decay
My name is Ann [436]
<span>From the message you sent me:

when you breathe normally, about 12 % of the air of your lungs is replaced with each breath. how much of the original 500 ml remains after 50 breaths

If you think of number of breaths that you take as a time measurement, you can model the amount of air from the first breath you take left in your lungs with the recursive function

b_n=0.12\times b_{n-1}

Why does this work? Initially, you start with 500 mL of air that you breathe in, so b_1=500\text{ mL}. After the second breath, you have 12% of the original air left in your lungs, or b_2=0.12\timesb_1=0.12\times500=60\text{ mL}. After the third breath, you have b_3=0.12\timesb_2=0.12\times60=7.2\text{ mL}, and so on.

You can find the amount of original air left in your lungs after n breaths by solving for b_n explicitly. This isn't too hard:

b_n=0.12b_{n-1}=0.12(0.12b_{n-2})=0.12^2b_{n-2}=0.12(0.12b_{n-3})=0.12^3b_{n-3}=\cdots

and so on. The pattern is such that you arrive at

b_n=0.12^{n-1}b_1

and so the amount of air remaining after 50 breaths is

b_{50}=0.12^{50-1}b_1=0.12^{49}\times500\approx3.7918\times10^{-43}

which is a very small number close to zero.</span>
5 0
3 years ago
HELP QUICK PLEASE 20 POINTS
trasher [3.6K]

Answer:

-2

Step-by-step explanation:

-5n + 18

where n is the number of the term


4 0
4 years ago
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