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3 years ago

Classify each pair of labeled angles of a complementary, supplementary, or neither.

Mathematics

1 answer:

Vsevolod [243]3 years ago

8 0

Answer:

The first one is neither: 90+89=179

The second one could be either supplementary or neither: 63+47=110, but if you add the final angle measure it would be 180

The last one is complementary: 62+29=90

<em>Hope this helps; Please mark Brainliest :)</em>

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600 or 600 decreased by 20% and then increased by 20% of the result? Which amount is larger and by how much

ohaa [14]

Answer:

The initial value of 600 is larger and larger by 24.

Step-by-step explanation:

First of all a 20% decreased of 600.

So, the new value will be $600(1 - \frac{20}{100}) = 600 \times 0.8 = 480$ .

Now, then it is increased by 20% i.e. 480 will be increased by 20%.

So, the new value will be $480(1 + \frac{20}{100}) = 480 \times 1.2 = 576$ .

Therefore, the initial value of 600 is larger and larger by (600 - 576) = 24. (Answer)

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2 years ago

to get a certain shade of green paint, Melissa decides to mix 3 gallons of blue paint with 2 gallons of yellow paint. How much b

maw [93]

Melissa will need 1.5 gallons of blue paint for each gallon of yellow paint.

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3 years ago

Help me!!!!!!!!!!!!!

irakobra [83]

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2 years ago

Read 2 more answers

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 assum

Ivan

Answer:

Differential equation

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

Solution

$y=\frac{1}{1+4e^{-0.327t}}$

Value of constant k=0.327 days^(-1)

The rumor reaches 80% at 8.48 days.

Step-by-step explanation:

We know

y(t): proportion of people that heard the rumor

y'(t)=ky(1-y), rate of spread of the rumor

Differential equation

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

Solving the differential equation

$\frac{dy}{y(1-y)}=k\cdot dt \\\\\int \frac{dx}{y(1-y)} =k \int dt \\\\-ln(1-\frac{1}{y} )+C_0=kt\\\\1-\frac{1}{y} =Ce^{-kt}\\\\\frac{1}{y} =1-Ce^{-kt}\\\\y=\frac{1}{1-Ce^{-kt}}$

Initial conditions:

$y(0)=0.2\\y(3)=0.4\\\\y(0)=0.2=\frac{1}{1-Ce^0}\\\\1-C=1/0.2\\\\C=1-1/0.2= -4\\\\\\y(3)=0.4=\frac{1}{1+4e^{-3k}} \\\\1+4e^{-3k}=1/0.4\\\\e^{-3k}=(2.5-1)/4=0.375\\\\k=ln(0.375)/(-3)=0.327\\\\\\y=\frac{1}{1+4e^{-0.327t}}$

Value of constant k=0.327 days^(-1)

At what time the rumor reaches 80%?

$y(t)=0.8=\frac{1}{1+4e^{-0.327t}} \\\\1+4e^{-0.327t}=1/0.8=1.25\\\\e^{-0.327t}=(1.25-1)/4=0.0625\\\\t=ln(0.0625)/(-0.327)=8.48$

The rumor reaches 80% at 8.48 days.

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3 years ago

In step 7, you listed three elements commonly listed in charts. what are all of those options? check all that apply. data points

Vanyuwa [196]

Data points
a data series
an x axis
a y axis
a legend

hope this helps (:

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2 years ago