Check the picture below.
let's firstly convert the mixed fractions to improper fractions.
![\stackrel{mixed}{7\frac{1}{2}}\implies \cfrac{7\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{15}{2}} ~\hfill \stackrel{mixed}{12\frac{1}{2}}\implies \cfrac{12\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{25}{2}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cstackrel%7Bmixed%7D%7B7%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B7%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B15%7D%7B2%7D%7D%20~%5Chfill%20%5Cstackrel%7Bmixed%7D%7B12%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B12%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B25%7D%7B2%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\stackrel{\textit{\Large Areas}}{\stackrel{two~triangles}{2\left[ \cfrac{1}{2}\left(\cfrac{15}{2} \right)(10) \right]}~~ + ~~\stackrel{\textit{three rectangles}}{(10)(15)~~ + ~~\left( \cfrac{15}{2} \right)(15)~~ + ~~\left( \cfrac{25}{2} \right)(15)}} \\\\\\ 75~~ + ~~150~~ + ~~112.5~~ + ~~187.5\implies \boxed{525}](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Ctextit%7B%5CLarge%20Areas%7D%7D%7B%5Cstackrel%7Btwo~triangles%7D%7B2%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%5Cleft%28%5Ccfrac%7B15%7D%7B2%7D%20%5Cright%29%2810%29%20%5Cright%5D%7D~~%20%2B%20~~%5Cstackrel%7B%5Ctextit%7Bthree%20rectangles%7D%7D%7B%2810%29%2815%29~~%20%2B%20~~%5Cleft%28%20%5Ccfrac%7B15%7D%7B2%7D%20%5Cright%29%2815%29~~%20%2B%20~~%5Cleft%28%20%5Ccfrac%7B25%7D%7B2%7D%20%5Cright%29%2815%29%7D%7D%20%5C%5C%5C%5C%5C%5C%2075~~%20%2B%20~~150~~%20%2B%20~~112.5~~%20%2B%20~~187.5%5Cimplies%20%5Cboxed%7B525%7D)
No. x cannot equal 2 and 7
Answer:
yes
Step-by-step explanation:
The line intersects each parabola in one point, so is tangent to both.
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For the first parabola, the point of intersection is ...
y^2 = 4(-y-1)
y^2 +4y +4 = 0
(y+2)^2 = 0
y = -2 . . . . . . . . one solution only
x = -(-2)-1 = 1
The point of intersection is (1, -2).
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For the second parabola, the equation is the same, but with x and y interchanged:
x^2 = 4(-x-1)
(x +2)^2 = 0
x = -2, y = 1 . . . . . one point of intersection only
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If the line is not parallel to the axis of symmetry, it is tangent if there is only one point of intersection. Here the line x+y+1=0 is tangent to both y^2=4x and x^2=4y.
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Another way to consider this is to look at the two parabolas as mirror images of each other across the line y=x. The given line is perpendicular to that line of reflection, so if it is tangent to one parabola, it is tangent to both.
C: 9 ( = 3^2), 49 ( = 7^2), 81 ( = 9^2)