HELP ME THIS IS DUE RN. ILL give 50 points and brainiest.

Mathematics

2 answers:

Setler [38]3 years ago

7 0

Answer:

y=1x+1

Step-by-step explanation:

it keeps going up one everytime

1x - 14y

2x - 15y

3x - 16y

4x - 17y

5x - 18y

6x - 19y

7x - 20y

8x - 21y

And so on

Maru [420]3 years ago

3 0

Answer:

y = 8x + 21

Step-by-step explanation:

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6 out of the 60 employees at the sports shop are part-time employees. What percentage of the employees at the sports shop work p

Serhud [2]

Percentage of part time employees = (part-time employees/total employees) * 100 = (6/60) * 100 = 1/10 * 100 = 10%

6 0

3 years ago

A rumor spreads through a small town. Let y(t) be the fraction of the population that has heard the rumor at time t and assume t

sladkih [1.3K]

Answer:

The answer is shown below

Step-by-step explanation:

Let y(t) be the fraction of the population that has heard the rumor at time t and assume that the rate at which the rumor spreads is proportional to the product of the fraction y of the population that has heard the rumor and the fraction 1−y that has not yet heard the rumor.

$\frac{dy}{dt}\ \alpha\ y(1-y)$

$\frac{dy}{dt}=ky(1-y)$

where k is the constant of proportionality, dy/dt = rate at which the rumor spreads

$\frac{dy}{dt}=ky(1-y)\\\frac{dy}{y(1-y)}=kdt\\\int\limits {\frac{dy}{y(1-y)}} \, =\int\limit {kdt}\\\int\limits {\frac{dy}{y}} +\int\limits {\frac{dy}{1-y}} =\int\limit {kdt}\\\\ln(y)-ln(1-y)=kt+c\\ln(\frac{y}{1-y}) =kt+c\\taking \ exponential \ of\ both \ sides\\\frac{y}{1-y} =e^{kt+c}\\\frac{y}{1-y} =e^{kt}e^c\\let\ A=e^c\\\frac{y}{1-y} =Ae^{kt}\\y=(1-y)Ae^{kt}\\y=\frac{Ae^{kt}}{1+Ae^{kt}} \\at \ t=0,y=10\%\\0.1=\frac{Ae^{k*0}}{1+Ae^{k*0}} \\0.1=\frac{A}{1+A} \\A=\frac{1}{9} \\$