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lina2011 [118]
3 years ago
6

A function f is described by f(x)=A*exp(kx)+B, where A, B and k are constants. Given f(0)=1, f(1)=2, and that the horizontal asy

mptote of f is -4, the value of k is
Mathematics
1 answer:
s2008m [1.1K]3 years ago
8 0

Answer:

k = ln (6/5)

Step-by-step explanation:

for

f(x)=A*exp(kx)+B

since f(0)=1, f(1)=2

f(0)= A*exp(k*0)+B = A+B = 1

f(1) = A*exp(k*1)+B =  A*e^k + B = 2

assuming k>0 , the horizontal asymptote H of f(x) is

H= limit f(x) , when x→ (-∞)

when x→ (-∞) , limit f(x) =  limit (A*exp(kx)+B) = A* limit [exp(kx)]+B* limit = A*0 + B = B

since

H= B = (-4)

then

A+B = 1 → A=1-B = 1 -(-4) = 5

then

A*e^k + B = 2

5*e^k + (-4) = 2

k = ln (6/5)    ,

then our assumption is right and k = ln (6/5)  

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ASAP please
Slav-nsk [51]

Answer:

umm i think you would be at tje point 4,0

3 0
3 years ago
Mel used partial quotients to find the quotient of 378 divided by 3.What could be the partial quotient that Mel found?
Genrish500 [490]

Answer:

The partial quotients are 100, 20 and 6. The quotient is 126.

Step-by-step explanation:

The dividend is 378 and the divisor is 3.

First find the factors of 3, near to 300.

3\times 1=3

3\times 10=30

3\times 100=300

The value 300 is less than 378, so record the partial quotient 100 and subtract 300 from 378. Repeat the same process until the dividend has been zero.

Now the remaining divide is 78.

First find the factors of 3, near to 78.

3\times 20=60

3\times 30=90

Since 90>78, therefore 30 is not a partial quotient. The value 60 is less than 78 , so record the partial quotient 20 and subtract 60 from 78.

Now the remaining divide is 18.

3\times 6=18

The number 6 is a partial quotient.

Therefore  partial quotients are 100, 20 and 6. Add the partial quotient to find quotient.

100+20+6=126

3 0
3 years ago
Compare these two functions, where the input variable represents days.
olga2289 [7]

Answer:

c.) 70

# carry on learning

4 0
3 years ago
(x – 2) / 4 – (3x + 5) / 7 = – 3, x = ?<br><br> WILL GIVE BRAINLIEST
balu736 [363]

Answer:

under

Step-by-step explanation:

change the underminator, we have

7×(x-2)-4×(3x+5)=-3×7×4

7x-14-12x-20=-84

-5x=-50

x=10

6 0
3 years ago
Based on the data in the two-way table, the probability of being 25-35 years and having a hemoglobin level above 11 is __ a.29 b
Alina [70]

The probability of being 25-35 years and having a haemoglobin level above 11 is 34%.

The probability of having a haemoglobin level above 11 is 36%.

Being 25-35 years and having a hemoglobin level above 11 are not dependent on each other.

<h3>What are the probabilities?
</h3>

Probability determines the odds that a random event would occur. The odds of the event happening lie between 0 and 1.

The probability of being 25-35 years and having a haemoglobin level above 11 = number of people between 25 - 35 that have a level above 11 / total number of people between 25 - 35

44 / 128 = 34%

The probability of having a haemoglobin level above 11  = number of people with a level above 11 / total number of respondents

153 / 429 = 36%

To learn more about probability, please check: brainly.com/question/13234031

#SPJ1

4 0
2 years ago
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