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2 years ago

On a coordinate plane, a circle has a center at point (negative 6, 4).

Mathematics

1 answer:

liraira [26]2 years ago

7 0

Answer:

${x}^{2} + {y}^{2} + 12x - 8y + 4 = 0$

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kotegsom [21]

Answer:

The Answer is D. If the month is on the x axis then its gonna be above the horazontal line (x-axis) and if its a positive temerature then its gonna be in the quadrent 1.

Step-by-step explanation:

8 0

2 years ago

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Solve the inequality 2n – 1 < 39

MatroZZZ [7]

Answer:

n < 20

Step-by-step explanation:

2n – 1 < 39

Add 1 to each side

2n – 1+1 < 39+1

2n < 40

Divide each side by 2

2n/2 < 40/2

n < 20

3 0

2 years ago

Use the power series for 1 1−x to find a power series representation of f(x) = ln(1−x). What is the radius of convergence? (Note

Viktor [21]

a. Recall that

$\displaystyle\int\frac{\mathrm dx}{1-x}=-\ln|1-x|+C$

For $|x|$ , we have

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

By integrating both sides, we get

$\displaystyle-\ln(1-x)=C+\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}$

If $x=0$ , then

$\displaystyle-\ln1=C+\sum_{n=0}^\infty\frac{0^{n+1}}{n+1}\implies 0=C+0\implies C=0$

so that

$\displaystyle\ln(1-x)=-\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}$

We can shift the index to simplify the sum slightly.

$\displaystyle\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}n$

b. The power series for $x\ln(1-x)$ can be obtained simply by multiplying both sides of the series above by $x$ .

$\displaystyle x\ln(1-x)=-\sum_{n=1}^\infty\frac{x^{n+1}}n$

c. We have

$\ln2=-\dfrac\ln12=-\ln\left(1-\dfrac12\right)$

$\displaystyle\implies\ln2=\sum_{n=1}^\infty\frac1{n2^n}$

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2 years ago

Please help!!!! I'm terrible at algebra and I need your help!!!

noname [10]

It will be the third one

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2 years ago

What property or properties are not used when solving the following equation -3(4x-8)=-36

BigorU [14]

Either B or D I’m not really sure but I know it’s one of this answers

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2 years ago

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