Simplify the expression.

Mathematics

1 answer:

levacccp [35]2 years ago

4 0

Hi there! The answer is 30.

We need to use PEMDAS to simplify.
P = parenthesis
E = Exponents
M = Multiply
D = Dividing
A = Adding
S = Subtracting

$10(4 \times 10 \div ( {6}^{2} - {4}^{2} ) + 1)$
First find the exponents inside the middlest parenthesis.

$10(4 \times 10 \div (36 - 16) + 1)$
Now add inside the parenthesis.

$10(4 \times 10 \div 20 + 1)$
Multiply inside the parenthesis that are left.

$10(40 \div 20 + 1)$
Divide inside the parenthesis

$10(2 + 1)$
Add inside the parenthesis.

$10(3)$
Multiply.

$30$
~ Hope this helps you!

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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

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