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3 years ago

In what quadrant will you find negative y and negative x

Mathematics

1 answer:

Molodets [167]3 years ago

6 0

<span>Second quadrant (II).
Hope that helps</span>

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A high school student named David Merrell did a study of how music affects the ability of rats to run a maze. He trained 71 rats

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3 years ago

What is an expression that compares two or more numbers

Kazeer [188]

A ratio comparing two quantities measures in different units . An equation stating that two ratios or rates are equivalent. Percent proportion . One ratio or function compares part of a quantity to the whole quantity.

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3 years ago

The cost of a jacket increased from $85.00 to $99.45. What is the percentage increase of the cost of the jacket?

trapecia [35]

Answer:

17%

Step-by-step explanation:

99.45-85=14.45

14.45/85= 0.17

0.17= 17%

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4 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

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

$y=\frac{\frac{1}{9} e^{kt}}{1+\frac{1}{9} e^{kt}}\\y=\frac{1}{1+9e^{-kt}}$

At t = 2, y = 40% = 0.4

c) At y = 75% = 0.75

$y=\frac{1}{1+9e^{-0.8959t}}\\0.75=\frac{1}{1+9e^{-0.8959t}}\\t=3.68\ days$

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3 years ago

Do you guys know the angle?

Paha777 [63]

90 degrees :).........

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3 years ago