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Tanya [424]
3 years ago
12

If f(x) = 3LX-2], what is f(5.9)? 9 10 11 12​

Mathematics
2 answers:
Black_prince [1.1K]3 years ago
6 0

Answer: 12

Step-by-step explanation:

We know that , the ceiling function y = [x] is also known as the least integer function that gives the smallest integer greater than or equal to x.

For example : For x= 1.5

y = [1.5] =2

For x= 3.64

y = ⌊3.64⌋=4

The given function :

Then, for x= 5.9 , we have

[since [3.9]=4 (least integer function)]

Therefore, the value of f(5.9) is 12

Illusion [34]3 years ago
3 0

d. <em>12</em>

<em>have a nice day <3</em>

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Scrat [10]

Volume of 1/4 gal

= Volume of 2/3 bin


Volume(x gal)

= Volume( 1 bin)


So, (1/4)/(2/3)

= x/1


(1/4)/(2/3)

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(1/4)/(3/2)

x = 3/8 (Total balance required to fill the entire bin)




In order to fill the bin, the additional gallons need to be added to the initial 1/4 gallon



Therefore, 3/8 - 1/4 = 1/8 gallons










Hope that helps!!!!



7 0
3 years ago
Which numbers are rational? Select all
Roman55 [17]

The answers would be A,D,E,F because ...

A rational number is any integer, fraction, terminating decimal, or repeating decimal.

Hope this helps! :3

5 0
3 years ago
Find the midpoint of the segment with the given endpoints <br><br> (-9,-9) and (7,5)
Mice21 [21]

Answer:

  • (-1, -2)

Step-by-step explanation:

<u>Given points</u>

  • (-9,-9) and (7,5)

<u>Midpoint is</u>

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3 years ago
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Troyanec [42]
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3 years ago
A small military base housing 1,000 troops, each of whom is susceptible to a certain virus infection. Assuming that during the c
slava [35]

Answer:

I=\frac{1000}{exp^{0,806725*t-0.6906755}+1}

Step-by-step explanation:

The rate of infection is jointly proportional to the number of infected troopers and the number of non-infected ones. It can be expressed as follows:

\frac{dI}{dt}=a*I*(1000-I)

Rearranging and integrating

\frac{dI}{dt}=a*I*(1000-I)\\\\\frac{dI}{I*(1000-I)}=a*dt\\\\\int\frac{dI}{I*(1000-I)}=\int a*dt\\\\-\frac{ln(1000/I-1)}{1000}+C=a*t

At the initial breakout (t=0) there was one trooper infected (I=1)

-\frac{ln(1000/1-1)}{1000}+C=0\\\\-0,006906755+C=0\\\\C=0,006906755

In two days (t=2) there were 5 troopers infected

-\frac{ln(1000/5-1)}{1000}+0,006906755=a*2\\\\-0,005293305+0,006906755=2*a\\a = 0,00161345 / 2 = 0,000806725

Rearranging, we can model the number of infected troops (I) as

-\frac{ln(1000/I-1)}{1000}+0,006906755=0,000806725*t\\\\-\frac{ln(1000/I-1)}{1000}=0,000806725*t-0,006906755\\-ln(1000/I-1)=0,806725*t-0.6906755\\\\\frac{1000}{I}-1=exp^{0,806725*t-0.6906755}  \\\\\frac{1000}{I}=exp^{0,806725*t-0.6906755}+1\\\\I=\frac{1000}{exp^{0,806725*t-0.6906755}+1}

6 0
3 years ago
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