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2 years ago

PLEASE HELP I NEED HELP ASAP

Mathematics

1 answer:

Slav-nsk [51]2 years ago

5 0

Answer:

$y=-x+15$

Step-by-step explanation:

Slope-intercept form is written as $y=mx+b$ where $m$ is the slope and $b$ is the y-intercept.

Any line perpendicular to a given line with slope $m$ has a slope of $-\frac{1}{m}$ . Therefore, the slope of a line perpendicular to $y=x+12$ is $-\frac{1}{1}=-1$ .

We can now plug the coordinate it passes through into our slope-intercept form equation to solve for the y-intercept:

$3=-1(12)+b,\\b=15$

Therefore, the equation of the line that is perpendicular to $y=x+12$ and passes through the coordinates $(12,3)$ is $\fbox{$y=-x+15$}$ .

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Step-by-step explanation:

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Hi there!

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We can also write is out entirely.
${x}^{2} = x \times x$
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Identify whether the series sigma notation infinity i=1 15(4)^i-1 is a convergent or divergent geometric series and find the sum

const2013 [10]

Answer: The correct option is

(d) This is a divergent geometric series. The sum cannot be found.

Step-by-step explanation: The given infinite geometric series is

$S=\sum_{i=1}^{\infty}15(4)^{i-1}.$

We are to identify whether the given geometric series is convergent or divergent. If convergent, we are to find the sum of the series.

We have the D' Alembert's ratio test, states as follows:

Let, $\sum_{i=1}^{\infty}a_i$ is an infinite series, with complex coefficients $a_i$ and we consider the following limit:

$L=\lim_{i\rightarrow \infty}\dfrac{a_{i+1}}{a_i}.$

Then, the series will be convergent if L < 1 and divergent if L > 1.

For the given series, we have

$a_i=15(4)^{i-1},\\\\a_{i+1}=15(4)^i.$

So, the limit is given by

$L\\\\\\=\lim_{i\rightarrow \infty}\dfrac{a_{i+1}}{a_i}\\\\\\=\lim_{i\rightarrow \infty}\dfrac{15(4)^i}{15(4)^{i-1}}\\\\\\=\lim_{i\rightarrow \infty}\dfrac{15(4)^i}{15(4)^{i}4^{-1}}\\\\\\=\dfrac{1}{4^{-1}}\\\\=4>1.$