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Ivenika [448]
3 years ago
11

It n is odd, which represents the first consecutive even inter after n

Mathematics
1 answer:
nikklg [1K]3 years ago
3 0
Let's pick an odd number hmmm say 7, 7 is odd.

what's the first consecutive integer after it?  well is 8, so is just one hop away from 7, or 7+1.

so for any odd integer, the next even is just one hop away.
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PLZ
Lena [83]

Answer:

$14.50 I think

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7 0
3 years ago
PLEASE HELP?!?!
Nonamiya [84]

Answer:

Dilation factor is 2/3

Step-by-step explanation:

Look at both flat lines (M to S and M' to S') and find their lengths.

M to S = 9 units

M' to S' = 6 units

The triangle has gotten smaller so the dilation factor is a fraction

new/original = 6/9 = .66666666 or 2/3

Dilation factor is 2/3

3 0
4 years ago
A concert sold out 12 perfomances.Altogether,8,208 tikets were sold.How many tickets were sold for each perfomance?
Sonbull [250]
8,208 divided by 12 = 684 ticket plz mark me as brainliest
HOPE THIS HELPS!!!!!!! ;-) 
3 0
3 years ago
Read 2 more answers
At time t hours after taking the cough suppressant hydrocodone bitartrate, the amount, A, in mg, remaining in the body is given
Nostrana [21]

Answer:

a. The initial amount was 10 mg.

b. The percentage of the drug leaving the body each hour is 0.18, this is 18% per hour.

c. The amount of drug that remains in the body 6 hours after dosing is 3.04 mg.

d.  The time until only 1 mg of the drug remains in the body is 11.6 hours.

Step-by-step explanation:

You know that at time t hours after taking the cough suppressant hydrocodone bitartrate, the amount, A, in mg, remaining in the body is given by :

A=10*(0.82)^{t}

a. The initial quantity occurs when time t is the initial t, that is, t is equal to 0. Then:

A=10*(0.82)^{t}=10*(0.82)^{0}=10*1\\

A=10

<u><em>The initial amount was 10 mg.</em></u>

b. Considering that an exponential growth is determined by:

A=A0*(1-r)^{t}, where A is the amount after a certain number of a certain time, Ao is the initial amount, r is the rate and t is the time, so the percentage of the drug that leaves the body each hour is :

1-r=0.82

Solving:

1-r -1= 0.82 -1

-r= -0.18

r= 0.18

<u><em>The percentage of the drug leaving the body each hour is 0.18, this is 18% per hour.</em></u>

c. The amount of drug that remains in the body 6 hours after dosing is when t = 6:

A=10*(0.82)^{6}

Solving:

A= 3.04 mg

<u><em>The amount of drug that remains in the body 6 hours after dosing is 3.04 mg.</em></u>

d. The time that passes until only 1 mg of the drug remains in the body is calculated taking into account that A = 1 mg:

1=10*(0.82)^{t}

Solving:

0.1=(0.82)^{t}

㏒ 0.1= t*㏒ 0.82

㏒ 0.1  ÷ ㏒ 0.82= t

11.6 hours= t

<u><em> The time until only 1 mg of the drug remains in the body is 11.6 hours.</em></u>

4 0
3 years ago
a construction crew is digging a hole. On the first day, they dug a hole 3 feet deep. On the second day, they dug 2 more feet. O
Tamiku [17]
(-3)+(-2)+(-4)= -9. The hole is 9 feet deep. It's related to the problem because it's the total depth of the hole.
3 0
4 years ago
Read 2 more answers
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