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Sholpan [36]
2 years ago
13

Find the radius of a circle that has this measurement. d=2.25cm

Mathematics
1 answer:
yuradex [85]2 years ago
6 0
I would say your answer would be  1.125 because the radius is half of the diameter so 2.25 / 2  gives you 1.125.
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Hi! please help i’ll give brainliest
Studentka2010 [4]

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A because the river is fastest at that shore

Step-by-step explanation:

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2 years ago
I just need help with 14,15,and the bonus
ira [324]

Answer:

9: a^8/c^2

Step-by-step explanation:

(a^4/c)^2

(a^8/c^2)

3 0
2 years ago
What is an equation of the line passing between(4, -6) and (12, -8)?*
I am Lyosha [343]

Answer:

y= -1/4x - 5

Step-by-step explanation:

first you need to find the slope.

to do this you can use the equation: change in y/change in x which is also y2 - y1/x2-x1

-8 - (-6)/12-4

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since the line's equation is y = mx + b, we still need to find the y-intercept. we can do this by just plugging in one of the points

-6 = -1/4(4) + b

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4 0
2 years ago
I WILL MARK BRAINIEST<br> I got one correct there should be about one more that’s also right
Lesechka [4]

So essentially an adjacent angle is when two angles have a common side and a common vertex.

So knowing that we can determine from the picture all the adjacent angles:

D) <2 and <3 are adjacent angles

E) <1 and <3 are adjacent angles

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I hope this helps.

7 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 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.

8 0
3 years ago
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