bearing in mind that perpendicular lines have negative reciprocal slopes, so
![\bf \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}~\hspace{10em}\stackrel{slope}{y=\stackrel{\downarrow }{-\cfrac{1}{3}}x-1} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20slope-intercept~form%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y%3D%5Cunderset%7By-intercept%7D%7B%5Cstackrel%7Bslope%5Cqquad%20%7D%7B%5Cstackrel%7B%5Cdownarrow%20%7D%7Bm%7Dx%2B%5Cunderset%7B%5Cuparrow%20%7D%7Bb%7D%7D%7D%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D~%5Chspace%7B10em%7D%5Cstackrel%7Bslope%7D%7By%3D%5Cstackrel%7B%5Cdownarrow%20%7D%7B-%5Ccfrac%7B1%7D%7B3%7D%7Dx-1%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
so we're really looking for a line whose slope is 3 and runs through (1,5)
The answer is true;
because if a triangle had more than one obtuse angle it would become like a rhombus or something else so it is not possible to remain a triangle and have more than one obtuse angle c:
The answer is the third choice
-9,-3
Answer:
You would solve this as if the inequality sign was an equal sign.
Just make sure when you divide by a negative or multiply by a negative the sign switches, anyways:
5 + 5(x +4) < 20
Distribute the 5 and simplify
5 + 5x + 20 < 20
25 + 5x < 20
Isolate x (subtract 25 from both sides)
5x < -5
Divide by 5
x < -1
Yes, This is true. If the asymptotes of a hyperbola are the x- and y- axes, it IS known as a rectangular hyperbola.
