Please help me with this

Find the distance between the parallel lines m and n whose equations are

MA_775_DIABLO [31]

Answer:Find the distance between the parallel lines m and n whose equations are y = x + 4 and y = x - 6, respectively.

There are several ways to do this...here's one

Let (0, 4) be a point on the first line

Then.......a line with a negative reciprocal slope going through this point will have the equation :

y = -x + 4........so......we can find the intersection of this line with y = x - 6....set both equations equal

-x + 4 = x - 6 add x, 6 to both sides

10 = 2x divide both sides by 2

5 = x

So...using -x + 4, the y value at intersection = -1.......

So...we just need to find the distance from (0,4) to ( 5, -1) =

√[ (5)^2 + (4 + 1)^2 ] = 5√2 ≈ 7.07 units

Here's a pic....AB is the distance with A = (0,4) and B = (5, -1)

Step-by-step explanation:

8 0

3 years ago

Find measures of each exterior angle of a regular polygon with 16 sides

Maksim231197 [3]

The measure of each exterior angle = the measure of each central angle.
each angle = 360 / 16 = 22.5

3 0

3 years ago

Read 2 more answers

How do I find the integral<br> ∫10(x−1)(x2+9)dxint10/((x-1)(x^2+9))dx ?

defon

$\int\frac{10}{(x-1)(x^2+9)}\ dx=(*)\\\\\frac{10}{(x-1)(x^2+9)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+9}=\frac{A(x^2+9)+(Bx+C)(x-1)}{(x-1)(x^2+9)}\\\\=\frac{Ax^2+9A+Bx^2-Bx+Cx-C}{(x-1)(x^2+9)}=\frac{(A+B)x^2+(-B+C)x+(9A-C)}{(x-1)(x^2+9)}\\\Updownarrow\\10=(A+B)x^2+(-B+C)x+(9A-C)\\\Updownarrow\\A+B=0\ and\ -B+C=0\ and\ 9A-C=10\\A=-B\ and\ C=B\to9(-B)-B=10\to-10B=10\to B=-1\\A=-(-1)=1\ and\ C=-1$

$(*)=\int\left(\frac{1}{x-1}+\frac{-x-1}{x^2+9}\right)\ dx=\int\left(\frac{1}{x-1}-\frac{x+1}{x^2+9}\right)\ dx\\\\=\int\frac{1}{x-1}\ dx-\int\frac{x+1}{x^2+9}\ dx=\int\frac{1}{x-1}-\int\frac{x}{x^2+9}\ dx-\int\frac{1}{x^2+9}\ dx=(**)\\\\\#1\ \int\frac{1}{x-1}\ dx\Rightarrow\left|\begin{array}{ccc}x-1=t\\dx=dt\end{array}\right|\Rightarrow\int\frac{1}{t}\ dt=lnt+C_1=ln(x-1)+C_1$

$\#2\ \int\frac{x}{x^2+9}\ dx\Rightarrow \left|\begin{array}{ccc}x^2+9=u\\2x\ dx=du\\x\ dx=\frac{1}{2}\ du\end{array}\right|\Rightarrow\int\left(\frac{1}{2}\cdot\frac{1}{u}\right)\ du=\frac{1}{2}\int\frac{1}{u}\ du\\\\\\=\frac{1}{2}ln(u)+C_2=\frac{1}{2}ln(x^2+9)+C_2$

$\#3\ \int\frac{1}{x^2+9}\ dx=\int\frac{1}{x^2+3^2}\ dx=\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3\\\\therefore:\\\\\#1;\ \#2;\ \#3\Rightarrow(**)=ln(x-1)+C_1-\frac{1}{2}ln(x^2+9)+C_2-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3$

$\boxed{=ln(x-1)-\frac{1}{2}ln(x^2+9)-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C}$

4 0

3 years ago

Y=x^2+12x+35 into the form y=(x-h)^2+k

LiRa [457]

Answer:

y = (x + 6)² - 1

Step-by-step explanation:

So there are two ways of doing it, completing the square or partial factoring (which is what i prefer and will show)

so we will get x first:

x^2+12x

We are factoring x

x(x+12)

Now you do x=-12/2

x=-6

Now that you have x, you plg that back into the equation to get y

y= -6^2 +12(-6) +35

= 36-36+35

=-1

Put into vertex form

y = (x + 6)² - 1

4 0

2 years ago

An angle measuring 468 n degrees Is in standard position. For which value of N will the terminal side fall on the axis

Fudgin [204]

N=5 I think, tell me if it is correct

6 0

3 years ago