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3 years ago

Help pls?? answer for me??f

Mathematics

1 answer:

Dafna11 [192]3 years ago

4 0

ANSWER

C PO SANA MAKATULONG PO YAN

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M is directly proportional to r squared

Nesterboy [21]

Answer:

m=84

r:m = 1:7

therefore when r= 1 then M=14

$r = 12 \\ m = 7 \times 12 \\ m = 84$

I think that's how you do it

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3 years ago

Expand:<br><img src="https://tex.z-dn.net/?f=f%28z%29%20%3D%20%20%5Cfrac%7B1%7D%7Bz%28z%20-%202%29%7D%20" id="TexFormula1" title

Monica [59]

Expand f(z) into partial fractions:

$\dfrac1{z(z-2)} = \dfrac12 \left(\dfrac1{z-2} - \dfrac1z\right)$

Recall that for |z| < 1, we have the power series

$\displaystyle \frac1{1-z} = \sum_{n=0}^\infty z^n$

Then for |z| > 2, or |1/(z/2)| = |2/z| < 1, we have

$\displaystyle \frac1{z-2} = \frac1z \frac1{1 - \frac2z} = \frac1z \sum_{n=0}^\infty \left(\frac 2z\right)^n = \sum_{n=0}^\infty \frac{2^n}{z^{n+1}}$

So the series expansion of f(z) for |z| > 2 is

$\displaystyle f(z) = \frac12 \left(\sum_{n=0}^\infty \frac{2^n}{z^{n+1}} - \frac1z\right)$

$\displaystyle f(z) = \frac12 \sum_{n=1}^\infty \frac{2^n}{z^{n+1}}$

$\displaystyle f(z) = \sum_{n=1}^\infty \frac{2^{n-1}}{z^{n+1}}$

$\displaystyle \boxed{f(z) = \frac14 \sum_{n=2}^\infty \frac{2^n}{z^n} = \frac1{z^2} + \frac2{z^3} + \frac4{z^4} + \cdots}$

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2 years ago

Find polynomial from zeros<br> Zeros 3i,2-i

SSSSS [86.1K]

Y=x4−4x3+14x2−36x+45

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3 years ago

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docker41 [41]

Answer:

-3/2 and 3/2

Step-by-step explanation:

y=4x^2-9

y=0

4x^2-9=0

4x^2=9

x^2=9/4

x1=-√9/4=-3/2

x2=√9/4=3/2

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4 years ago

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See photo hope it helps.

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