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sergij07 [2.7K]
3 years ago
5

May you help me with this question? :3

Mathematics
1 answer:
Sauron [17]3 years ago
8 0

Answer:

The answer is C. 1/2^4

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The diameter of an insect cell is 1.7 × 10–11 meter. Ariel wants to find the length of the radius of the cell. If she performs t
Juli2301 [7.4K]

Answer:

8.5E-12

Step-by-step explanation:

According to the Question,

Given That, The diameter of an insect cell is 1.7 × 10^-11 meters.

Ariel wants to find the length of the radius of the cell. Radius Is Half of the Diameter.

  • Thus, R = D/2   ⇒  (1.7*10^-11)/2 = 0.85× 10^-11
  • If she performs the calculations on her calculator is 8.5E-12.
6 0
2 years ago
Saul earned $25 for painting his neighbor's fence. He spent $11 at the store and put the rest into his bank account.
PtichkaEL [24]
The answer is $14 Saul put into his bank accout
3 0
3 years ago
Neeeeeed heeelp plz!!!
Blababa [14]
                             |
                             |__x_- 4___________              2x³/2x² =x
   2x² + 2x +3        | 2x³ - 6x² +7x +3
                            - (<span>2x³ +2x² +3x)
</span>                                      -8x² +4x  +3                -8x²/2x² = -4
                                    <span>-(-8x² -8x  -12)
</span>                                              12x +15

(x-4) + (12x +15)/(2x² + 2x +3) 
6 0
3 years ago
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
What is the solution set of the following quadratic equation?
Brut [27]
The answer is AAAAAAAAAAAAAAAA
4 0
3 years ago
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