What is the answer to this

Can somebody help me with these to questions how do you find the area of the circle

Grace [21]

13)

Circumference (C) = 2π·r

11π = 2π·r

11/2 = r

******************************

Area (A) = π·r²

= π·(11/2)²

= 121π/4

≈ 94.985 in²

14)

diameter = 2r

42 = 2r

21 = r

*******************

Area (A) = π·r²

= π·(21)²

= 441π

≈ 1384.74 in²

***************************

$280/1384.74 in²

x = $0.20

Answer: about 20 cents per square inch

6 0

3 years ago

A. how does your equation from 1b relate to the pendulum equation?'

Law Incorporation [45]

What is the equation ? Is there a picture to the question.

8 0

3 years ago

Help me plssssssssssssssssss

meriva

Answer:

kmkioppppppppoiiu yw quip all cm ms all pure 100.00

3 0

2 years ago

At time t=0 sec, a tank contains 15 oz of salt dissolved in 50 gallons of water. Then brine containing 88oz of salt per gallon o

maria [59]

Answer:

a) $s(t) =400- 385 e^{-\frac{1}{10} t}$

b) $s(t=25min) =400- 385 e^{-\frac{1}{10}25}=368.397$

Step-by-step explanation:

Part a

Assuming that the original concentration of salt is 8 oz/gal and that the rate of in is equal to the rate out = 5 gal/min.

For this case we know that the rate of change can be expressed on this way:

$Rate change = In-Out$

And we can name the rate of change as $\frac{ds}{dt}=rate change$

And our variable s would represent the amount of salt for any time t.

We know that the brine containing 8oz/gal and the rate in is equal to 5 gal/min and this value is equal to the rate out.

For the concentration out we can assume that is $\frac{s}{50gal}$

And now we can find the expression for the amount of salt after time t like this:

$\frac{dS}{dt}= 8 \frac{oz}{gal}(5\frac{gal}{min}) -\frac{s}{50gal} 5 \frac{gal}{min} =40\frac{oz}{min}- \frac{s}{10} \frac{oz}{min}$

And we have this differential equation:

$\frac{dS}{dt} +\frac{1}{10} s = 40$

With the initial conditions y(0)=15 oz

As we can see we have a linear differential equation so in order to solve it we need to find first the integrating factor given by:

$\mu = e^{\int \frac{1}{10} dt }= e^{\frac{1}{10} t}$

And then in order to solve the differential equation we need to multiply with the integrating factor like this:

$e^{\frac{1}{10} t} s = \int 40 e^{\frac{1}{10} t} dt$

$e^{\frac{1}{10} t} s = 400 e^{\frac{1}{10} t} +C$

Now we can divide both sides by $e^{\frac{1}{25} t}$ and we got:

$s(t) =400 + C e^{-\frac{1}{10} t}$

Now we can apply the initial condition in order to solve for the constant C like this:

$15 = 400+C$

$C=-385$

And then our function would be given by:

$s(t) =400- 385 e^{-\frac{1}{10} t}$

Part b

For this case we just need to replace t =25 and see what we got for the value of the concentration:

$s(t=25min) =400- 385 e^{-\frac{1}{10}25}=368.397$

7 0

2 years ago

Pedro has a balance of $5810 on a credit card with an APR of 16.2%. Paying off his balance in which of these lengths of time wil

Sonja [21]

A interest payed 399.19
B interest payed 481.12
C interest payed 563.76
D interest payed 317.95
you will pay the lest with a 7 month contract but you will pay more monthly but pay a lesser amount in the end so your answer is D

3 0

3 years ago