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kolezko [41]
3 years ago
8

PLEASE ANSWER + BRAINLIEST !!!

Mathematics
1 answer:
Jobisdone [24]3 years ago
7 0

Answer:


Step-by-step explanation:

12 i think

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Fractions and decimals
8 0
3 years ago
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9 less than the product of 37 and x
zepelin [54]
Alrighty, so, 9 less hints to subtract. The product of 37 and x means that you're multiplying 37 and x. Therefore, I believe the answer that you're looking for is 37x-9.

Hope this helps! (:
8 0
3 years ago
Find the radius of convergence of the power series. (if you need to use or –, enter infinity or –infinity, respectively. ) [infi
Alona [7]

For given power series \sum_{n=0}^{\infty} \frac{(-1)^nx^n}{2^n} the radius of convergence is 2.

For given question,

We have been given a power series \sum_{n=0}^{\infty} \frac{(-1)^nx^n}{2^n}

We need to find the radius of convergence of the power series.

We use ratio test to find the radius of convergence of the power series.

Let a_n=\frac{(-1)^nx^n}{2^n}

\Rightarrow a_{n+1}=\frac{(-1)^{n+1}x^{n+1}}{2^{n+1}}

Consider,

\lim_{n \to \infty}|\frac{a_{n+1}}{a_n} |\\\\= \lim_{n \to \infty} |\frac{\frac{(-1)^{n+1}x^{n+1}}{2^{n+1}}}{ \frac{(-1)^nx^n}{2^n} } |\\\\=\lim_{n \to \infty} |\frac{(-1)^{n+1}x^{n+1}}{2^{n+1}}\times \frac{2^n}{(-1)^nx^n} |\\\\=\lim_{n \to \infty} |\frac{(-1)x}{2} |\\\\=\lim_{n \to \infty}|\frac{-x}{2} |\\\\=\frac{x}{2}

By Ratio test, given power series converges at |\frac{x}{2} | < 1

⇒ |x| < 2

So, the radius of convergence is 2.

Therefore, for given power series \sum_{n=0}^{\infty} \frac{(-1)^nx^n}{2^n} the radius of convergence is 2.

Learn more about the radius of convergence here:

brainly.com/question/2289050

#SPJ4

8 0
2 years ago
Hi guys.
Veseljchak [2.6K]

Answer:

circle and rectangle

Step-by-step explanation:

4 0
3 years ago
Determine the next step for solving the quadratic equation by completing the square.
nexus9112 [7]

\qquad \textit{perfect square trinomial} \\\\ (a\pm b)^2\implies a^2\pm \stackrel{\stackrel{\text{\small 2}\cdot \sqrt{\textit{\small a}^2}\cdot \sqrt{\textit{\small b}^2}}{\downarrow }}{2ab} + b^2

the idea behind the completion of the square is simply using a perfect square trinomial,  hmmm usually we do that by using our very good friend Mr Zero, 0.

if we look at the 2nd step, we have a group as x² - x, hmmm so we need a third element, which will be squared.

keeping in mind that the middle term of the perfect square trinomial is simply the product of the roots of "a" and "b",  so in this case the middle term is "-x", and the 1st term is x², so we can say that

\stackrel{middle~term}{2(\sqrt{x^2})(\sqrt{b^2})~}~ = ~~\stackrel{middle~term}{-x}\implies 2xb~~ = ~~~~ = ~~-x \\\\\\ b=\cfrac{-x}{2x} \implies b=-\cfrac{1}{2}

so that means that our missing third term for a perfect square trinomial is simply 1/2, now we'll go to our good friend Mr Zero, if we add (1/2)², we have to also subtract (1/2)², because all we're really doing is borrowing from Zero, so we'll be including then +(1/2)² and -(1/2)², keeping in mind that 1/4 - 1/4 = 0, so let's do that.

-3~~ = ~~-2\left[ x^2-x+\left( \cfrac{1}{2} \right)^2 ~~ - ~~\left( \cfrac{1}{2} \right)^2\right]\implies -3=-2\left(x^2-x+\cfrac{1}{4}-\cfrac{1}{4} \right) \\\\\\ -3=-2\left(x^2-x+\cfrac{1}{4} \right)+(-2)-\cfrac{1}{4}\implies -3=-2\left(x^2-x+\cfrac{1}{4} \right)+\cfrac{1}{2} \\\\\\ -3-\cfrac{1}{2}=-2\left(x^2-x+\cfrac{1}{4} \right)\implies -\cfrac{7}{2}=-2\left(x-\cfrac{1}{2} \right)^2\implies \cfrac{7}{4}=\left(x-\cfrac{1}{2} \right)^2

~\dotfill\\\\ \pm\sqrt{\cfrac{7}{4}}=x-\cfrac{1}{2}\implies \cfrac{\pm\sqrt{7}}{2}=x-\cfrac{1}{2}\implies \cfrac{\pm\sqrt{7}}{2}+\cfrac{1}{2}=x \implies \cfrac{\pm\sqrt{7}+1}{2}=x

8 0
2 years ago
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