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Ksenya-84 [330]

3 years ago

6

The vector y = ai + bj is perpendicular to the line ax + by = c. Use this fact to find an equation of the line through P perpend

icular to v. Then draw a sketch of the line including v as a vector starting at the origin. P(-1,-9)v = -4i + j The equation of the line is _____.

1 answer:

PIT_PIT [208]3 years ago

6 0

Answer:

$-4x+y=-5$

Step-by-step explanation:

It is given that he vector y = ai + bj is perpendicular to the line ax + by = c.

Given given point is P(-1,-9) and given vector is v = -4i + j.

Here, a = -4 and b= 1.

We need to find the equation of the line through P perpendicular to v.

Using the above fact, the vector v = -4i + j is perpendicular to the line

$(-4)x+(1)y=c$

$-4x+y=c$ ... (1)

The line passes through the point (-1,-9).

Substitute x=-1 and y=-9 in the above equation to find the value of c.

$-4(-1)+(-9)=c$

$4-9=c$

$-5=c$

The value of c is -5.

Substitute c=-5 in equation (1).

$-4x+y=-5$

Therefore the equation of line which passes through P and perpendicular to v is -4x+y=-5.

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Evaluate the integral of the quantity x divided by the quantity x to the fourth plus sixteen, dx . (2 points) one eighth times t

Anika [276]

Answer:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

Step-by-step explanation:

Given

$\int\limits {\frac{x}{x^4 + 16}} \, dx$

Required

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$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

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Substitute $16u^2$ for $x^4$ in $\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16u^2 + 16}} \,\ du$

Simplify

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16}* \frac{1}{8u^2 + 8}} \,\ du$

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$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

In standard integration

$\int\limits {\frac{1}{u^2 + 1}} \,\ du = arctan(u)$

So, the expression becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(u)$

Recall that: $u = \frac{x^2}{4}$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

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