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3 years ago

PLEASE HELP!!! Write the equation of a line that goes through the point (6, 3) and is perpendicular to the line y = 4x + 1.

Mathematics

1 answer:

Tpy6a [65]3 years ago

8 0

For this case we have that by definition, the equation of a line in the point-slope form is given by:

$y-y_ {0} = m (x-x_ {0})$

Where:

m: It is the slope of the line and $(x_ {0}, y_ {0})$ is a point through which the line passes.

We have the following equation of the slope-intersection form:

$y = 4x + 1$

Where the slope is $m = 4$

By definition, if two lines are perpendicular then the product of their slopes is -1.

Thus, a perpendicular line will have a slope:

$m_ {2} = \frac {-1} {m_ {1}}\\m_ {2} = \frac {-1} {4}\\m_ {2} = - \frac {1} {4}$

Thus, the equation will be of the form:

$y-y_ {0} = - \frac {1} {4} (x-x_ {0})$

Finally we substitute the given point and we have:

$y-3 = - \frac {1} {4} (x-6)$

Answer:

Option B

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3 years ago

Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

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