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Ksenya-84 [330]
3 years ago
11

I need help with this!

Mathematics
1 answer:
Leviafan [203]3 years ago
7 0

Solution

f(r) = 3.14r^{2}

Now we have to find the area of the circle when the radius (r) = 4.

Plug in r = 4 in f(r) to get the area of the circle.

f(4) = 3.14(4^{2} )

f(4) = 3.14 * 4 * 4

f(4) = 3.14 *16

f(4) = 50.24

The answer is C. 50.24

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A sample of an unknown liquid has a volume of 24.0 mL and a mass of 6 g. What is its density?
Furkat [3]

Answer:

Hey there!

Density = Mass/Volume

Mass= 6g

Volume=24 mL, which is 24 cm^3

Density= 6g/24 cm^3

Density=0.25 g/cm^3

Let me know if this helps :)

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3 years ago
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When a bakery opened this morning the ratio of pineapple muffins to the number of mango muffins was 3:2. After 144 pineapple muf
Elena L [17]
3x-\ pineapple\\
2x-\ mango\\\\
after\ selling\ 144\ pineapple:\\
3y-\ pineapple\\
5y-\ mango\\\\
3x-144=3y\ \ | add\ 144\\
2x=5y\\\\
3x=3y+144\ \ \ | divide\ by\ 3\\
2x=5y\\\\
x=y+48\\2x=5y\\\\
2(y+48)=5y\\
2y+96=5y\ \ \ | subtract\ 2y\\
96=3y\ \ \ | divide\ by\ 3\\
32=y\\
x=32+48=80\\\\
There\ were\ 240\ pineapple\ muffins.

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3 years ago
Was the difference in the overall men's winner in the ages 70-74 group more less than 2 hours
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1) Use power series to find the series solution to the differential equation y'+2y = 0 PLEASE SHOW ALL YOUR WORK, OR RISK LOSING
iogann1982 [59]

If

y=\displaystyle\sum_{n=0}^\infty a_nx^n

then

y'=\displaystyle\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty(n+1)a_{n+1}x^n

The ODE in terms of these series is

\displaystyle\sum_{n=0}^\infty(n+1)a_{n+1}x^n+2\sum_{n=0}^\infty a_nx^n=0

\displaystyle\sum_{n=0}^\infty\bigg(a_{n+1}+2a_n\bigg)x^n=0

\implies\begin{cases}a_0=y(0)\\(n+1)a_{n+1}=-2a_n&\text{for }n\ge0\end{cases}

We can solve the recurrence exactly by substitution:

a_{n+1}=-\dfrac2{n+1}a_n=\dfrac{2^2}{(n+1)n}a_{n-1}=-\dfrac{2^3}{(n+1)n(n-1)}a_{n-2}=\cdots=\dfrac{(-2)^{n+1}}{(n+1)!}a_0

\implies a_n=\dfrac{(-2)^n}{n!}a_0

So the ODE has solution

y(x)=\displaystyle a_0\sum_{n=0}^\infty\frac{(-2x)^n}{n!}

which you may recognize as the power series of the exponential function. Then

\boxed{y(x)=a_0e^{-2x}}

7 0
3 years ago
5j+s=t-2 solve for t
Gekata [30.6K]

The answer is

T= 5j+s+2


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