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2 years ago

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Mathematics

2 answers:

Nikitich [7]2 years ago

5 0

The answer would be the first one :)

oksano4ka [1.4K]2 years ago

5 0

Answer:

answer is

slope=-2

y intercept=2

down 2 over 1 for slope

and the line meets the y axis at 2

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What is the sum of the geometric series

lbvjy [14]

If $S$ denotes the sum of the first $n$ terms of a geometric series with first term $a$ and common ratio $r$ , then

$S=a+ar+ar^2+\cdots+ar^{n-3}+ar^{n-2}+ar^{n-1}$
$rS=ar+ar^2+ar^3+\cdots+ar^{n-2}+ar^{n-1}+ar^n$
$\implies S-rS=a+(ar-ar)+(ar^2-ar^2)+\cdots+(ar^{n-1}-ar^{n-1})-ar^n$
$\implies (1-r)S=a(1-r^n)$
$\implies S=a=\dfrac{1-r^n}{1-r}$

Using summation notation, you have

$S=\displaystyle\sum_{x=1}^nar^{x-1}=\sum_{x=0}^{n-1}ar^x=a\dfrac{1-r^n}{1-r}$

In this case, you have $a=2$ , $r=\dfrac12$ , and $n=16$ . So the value of the sum is

$\displaystyle\sum_{x=0}^{15}2\left(\frac14\right)^x=2\dfrac{1-\left(\frac14\right)^{16}}{1-\frac14}\approx2.67$

Rounded to the nearest whole number, the answer would be 3.

4 0

3 years ago

Read 2 more answers

What are the roots of f(x) = x2 – 48? –48 and 48 –24 and 24 Negative 8 StartRoot 3 EndRoot and 8 StartRoot 3 EndRoot Negative 4

Ivanshal [37]

Answer:

$x=4\sqrt{3},\:x=-4\sqrt{3}$ are the roots.

Step-by-step explanation:

$x=4\sqrt{3},\:x=-4\sqrt{3}$

Considering the expression

$x^2\:-\:48$

Solving

$x^2-48=0$

$x^2-48+48=0+48$

$x^2=48$

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=\sqrt{48},\:x=-\sqrt{48}$

Solving

$x=\sqrt{48}$

= $\sqrt{2^4\cdot \:3}$

$\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$

= $\sqrt{3}\sqrt{2^4}$

$\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a^m}=a^{\frac{m}{n}}$

$\sqrt{2^4}=2^{\frac{4}{2}}=2^2$

$=2^2\sqrt{3}$

$=4\sqrt{3}$

So,

$x=4\sqrt{3}$

Similarly,

$x=-\sqrt{48}=-4\sqrt{3}$

Therefore, $x=4\sqrt{3},\:x=-4\sqrt{3}$ are the roots.

Keywords: roots, expression

Learn more about roots from brainly.com/question/3731376

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larisa [96]

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avanturin [10]

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