Answer: 7a^2+2a+7
Step-by-step explanation:Distribute the Negative Sign:
=9a2+7a+8+−1(2a2+5a+1)
=9a2+7a+8+−1(2a2)+−1(5a)+(−1)(1)
=9a2+7a+8+−2a2+−5a+−1
Combine Like Terms:
=9a2+7a+8+−2a2+−5a+−1
=(9a2+−2a2)+(7a+−5a)+(8+−1)
=7a2+2a+7
Answer:
Step-by-step explanation:
![x+y=k\implies y=k-x](https://tex.z-dn.net/?f=x%2By%3Dk%5Cimplies%20y%3Dk-x)
Substitute this into the parabolic equation,
![k-x=6-(x-3)^2\implies x^2-7x+(k+3)=0](https://tex.z-dn.net/?f=k-x%3D6-%28x-3%29%5E2%5Cimplies%20x%5E2-7x%2B%28k%2B3%29%3D0)
We're told the line
intersects
twice, which means the quadratic above has two distinct real solutions. Its discriminant must then be positive, so we know
![(-7)^2-4(k+3)=49-4(k+3)>0\implies k](https://tex.z-dn.net/?f=%28-7%29%5E2-4%28k%2B3%29%3D49-4%28k%2B3%29%3E0%5Cimplies%20k%3C%5Cdfrac%7B37%7D4%3D9.25)
We can tell from the quadratic equation that
has its vertex at the point (3, 6). Also, note that
![-1\le\sin t\le1\implies3\le3+2\sin t\le5\implies3\le x\le5](https://tex.z-dn.net/?f=-1%5Cle%5Csin%20t%5Cle1%5Cimplies3%5Cle3%2B2%5Csin%20t%5Cle5%5Cimplies3%5Cle%20x%5Cle5)
and
![-1\le\cos2t\le1\implies2\le4+2\cos2t\le6\implies2\le y\le6](https://tex.z-dn.net/?f=-1%5Cle%5Ccos2t%5Cle1%5Cimplies2%5Cle4%2B2%5Ccos2t%5Cle6%5Cimplies2%5Cle%20y%5Cle6)
so the furthest to the right that
extends is the point (5, 2). The line
passes through this point for
. For any value of
, the line
passes through
either only once, or not at all.
So
; in set notation,
![\{k\mid 7\le k](https://tex.z-dn.net/?f=%5C%7Bk%5Cmid%207%5Cle%20k%3C9.25%5C%7D)
Answer:
a ≈ 1.8
Step-by-step explanation:
a / sin (180 - 105 - 15)° = 2 /sin 105°
a = (2 /sin 105°) x sin 60°
a = (2 / 0.97) x 0.87
a = 1.79 (≈ 1.8)