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Line l1 represents the graph of 3x + 4y = -14. Line l2 passes through the point (-5,7), and is perpendicular to line l1. If line l2 represents the graph of y=mx +b, then find m+b.

first of all lets find its perp line 

$3x + 4y = -14 \implies 4x-3y=$perp line where given the condition that that perp line should go through $(-5,7)$, respectively $x=-5$ and $y=7$:

$4\left(-5\right)-3\left(7\right)=x$
$-20-21=x$
$-41=x$

thus, the perp line which goes through $(-5,7)$ is $4x-3y=-41$ BUT the thing is we want that in the form of $y=mx+b$

knowing what we have is in the form of $Ax + By = C$ to find $m$, the slope, we have to do $\frac{A}{B}$ so $\frac{4}{3} $

the y intercep, or $b$ is $\frac{-C}{B}$ so $\frac{-(-41)}{3} \implies\frac{41}{3}  $

so the line in $y=mx+b$ form is $ y=\frac{4}{3}x+\frac{41}{3} $ 

Finally, lets find $m+b$:

$\frac{4}{3}+\frac{41}{3} =\boxed{\frac{45}{3}} \Rightarrow \boxed{15} $

here is an image of it as well:

you couls mathematically solve them, or just use desmos

for $y = \frac{1}{3}x + 7$ the x value is $-21$ and for $ y = 2x+5$ the x value is $-2.5$

*What is the sum of the x-coordinates of the two lines' x-intercepts?*

$-21+-2.5=\boxed{-23.5}$

$ 2x^2+6x+11  $

$ 2\left(x^2+3x+\frac{11}{2}\right)  $

$ 2n=3 \implies n=\frac{3}{2} $ which you want to square $ \implies \left(\frac{3}{2}\right)^2 $

$ 2\left(x^2+3x+\frac{11}{2}+\left(\frac{3}{2}\right)^2\right) =2 \left( \left(\frac{3}{2}\right)^2 \right) $

$ 2\left(x^2+3x+\frac{11}{2}+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2\right)  $

completing the square 

$ 2\left(\left(x+\frac{3}{2}\right)^2+\frac{11}{2}-\left(\frac{3}{2}\right)^2\right)   \implies  2\left(\left(x+\frac{3}{2}\right)^2\right) + 2\left( \frac{11}{2}-\frac{9}{4} \right) $

           $\Updownarrow   $

$ 2\left(x+\frac{3}{2}\right)^2+\frac{26}{4} \equiv a(x - h)^2 + k $ 

$ \overset{. \: .}{\smile } $

are you asking for confirmation i dont know?

we have $9x+x=180 \therefore  x=18$

since 18 is the exterior angle, it means the interior angle is 162.

that gives you $ 162n\:=\:180n\:−\:360\: $which implies that $ n=20 $ so you are correct $\overset{. \: .}{\smile}  $

what you are given is $ \{ 4x + 5y < 200 \} $

now, how many pairs of coordinates are there to satisfy that, and also $ ∈ \mathbb{Z}  $?

a graph $ \{ 4x + 5y = 200 \} $ -- you will see a line -- the coordinates we wont need are the ones which are larger than 200, so we just expunge the top part, where the number gets higher right?

(NOTE: 0 is not a positive number)

check the image 

do not forget to include the coordinates where the purple line is on as well

at this point just count them all if you want an answer -- unless someones manage to find an easier way LOL

also, do not count the ones where the line intersects, as those would make it $=200$  $ \overset{. \: .}{\smile}  $

this is a dependant probability thing

so what we have is  A A A A 2 2 2 2 3 3 3 3 -- what is the chance of us getting all distinct letters?

eachtime we pull, we take one right card, and then the probability to find the other ones is by eliminating the left three of that particular card, so:

$ \frac{12}{12} \frac{8}{11} \frac{4}{10}   $

therefore, the prob. of us pulling an ace, a two and a three is $ \frac{16}{55}  $

well, we both are talking about different things -- though yeah it is 120x^3 as its in the power of 3, not two -- good catch 👍:D

wait actually, no, i do not think thats right.

the cube root and the exponential thingie do not just cancel -- it would be $ 3\sqrt{3}  $ no?

ahh i do not really remember how to work these, but i will do what i can

we are asked to find the coefficent of $ x^3  $ in $ \left(x+2\sqrt{3}\right)^5 $

in these types of binomials, your exponent power value is going to be the $n$ and the $s$, or $i$, or $a$, i do not know what you know them as is going to be the coefficient thinge -- the one we are looking for is $i=3$

so we know that $n=5$ and $i=3$

plug it all into $ \frac{n!}{i!\left(n-i\right)!}x^{\left(n-i\right)}\left(2\sqrt{3}\right)^i  $

as following, we work it out:

 $\frac{5!}{3!\left(5-3\right)!}x^2\left(2\sqrt{3}\right)^3  $

$\frac{120}{6\cdot \:2}x^2\left(2\sqrt{3}\right)^3  $

$ 10\left(2\sqrt{3}\right)^3x^2  $

$ 10\cdot \: 8\cdot \: 3^{\frac{1}{2}\cdot \:3} x^2  $ 

$  10\cdot \: 8\cdot \: 3^{\frac{3}{2}}  x^2    $

$    10\cdot \: 8\cdot \: 3^{\frac{3}{2}}  x^2    $

$ 10\cdot \:8\cdot \:3\sqrt{3}\cdot \: x^2  $

$ \boxed{240\cdot \sqrt{3}\cdot \: x^2 }  $

if you want to get into decimals it would be:

$  240\cdot 1.73\cdot \: x^2   $    

$\boxed{ 415.2\cdot \:x^2 }  $

:D

oh lord i never noticed -- yes they indeed do add up to 360

at that moment i was working with triangles so thats why i was thinking of 180 LOL