7) y= x^2-4x
8) y= 2x^2-3x+1
9) y= 9x^2-4
Answer:
D
Step-by-step explanation:
Given the 2 equations
4x + 8y = 21 → (1)
y = - 
x + 3 → (2)
Substitute y = - x + 3 into (1)
4x + 8(- x + 3 ) = 21 ← distribute parenthesis on left side
4x - 4x + 24 = 21 ( subtract 24 from both sides )
0 = - 3 ← not possible
This indicates the system has no solutions
![\stackrel{\textit{\LARGE Line A}}{(\stackrel{x_1}{-8}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{-5}~,~\stackrel{y_2}{4})} ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{4}-\stackrel{y1}{5}}}{\underset{run} {\underset{x_2}{-5}-\underset{x_1}{(-8)}}} \implies \cfrac{4 -5}{-5 +8}\implies -\cfrac{1}{3} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Ctextit%7B%5CLARGE%20Line%20A%7D%7D%7B%28%5Cstackrel%7Bx_1%7D%7B-8%7D~%2C~%5Cstackrel%7By_1%7D%7B5%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B-5%7D~%2C~%5Cstackrel%7By_2%7D%7B4%7D%29%7D%20~%5Chfill%20%5Cstackrel%7Bslope%7D%7Bm%7D%5Cimplies%20%5Ccfrac%7B%5Cstackrel%7Brise%7D%20%7B%5Cstackrel%7By_2%7D%7B4%7D-%5Cstackrel%7By1%7D%7B5%7D%7D%7D%7B%5Cunderset%7Brun%7D%20%7B%5Cunderset%7Bx_2%7D%7B-5%7D-%5Cunderset%7Bx_1%7D%7B%28-8%29%7D%7D%7D%20%5Cimplies%20%5Ccfrac%7B4%20-5%7D%7B-5%20%2B8%7D%5Cimplies%20-%5Ccfrac%7B1%7D%7B3%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
keeping in mind that perpendicular lines have negative reciprocal slopes, and that parallel lines have equal slopes, well, those two slopes above aren't either, so since they're neither, and they're different, that means that lines A and B intersect.
Answer:
Translation 4 units to the left., followed by
a translation 1 unit down.
Step-by-step explanation:
The parent function is y = x^2 which is a parabola that opens upwards and has a vertex at the point (0,0).
Y = (x + 4)^2 is the graph of x^2 translated 4 units to the left.
The - 1 translates the graph down 1 unit.
So the vertex of the new graph is at (-4, 1).
We either need to see a picture of this and/or get more information about the measurements of the triangle. In general, the area outside of the triangle will be the area of the semi-circle minus the area of the triangle itself, or: 1/2*49*3.14 - 1/2 b*h of the triangle. That first part, which is the area of the semi-circle, works out to 76.93 So based on the info we have, it becomes 76.93 - 1/2*b*h of the triangle = area outside of triangle.
