Y = -0.4x
1) It is a straight line
2) I passes through the origin (0,0), because the y-intercpet is 0.
3) The slope is negative, so it passes throuh II and III quadrants
4) The magnitude of the slope = 0.4
4) The angle of the line with the negative side of the x-axis is that whose tan is 0.4 => angle = 21.8 °
With all that information you can identify the graph, given that you didn't include the options.
Answer:
The values for both of the missing sides is 8.
Step-by-step explanation:
The hypotenuse of a right triangle = a(square root of 2)
Therefore the other two missing sides equal a.
Therefore, the other two sides equal 8.
4 because 2cups of 1/2 cups=1cup
2x2=4
We have a pair of vertical angles here. So they would equal each other.
3x+50=6x-10
Subtract 3x to 6x:
50=3x-10
Add 10 to 50:
60=3x
Divide 3 to both sides:
x=30
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)
