we know the circle's center is at -7, -1, and we know the circle itself passes through 8,7, the distance from the center to a point on it is by definition its radius, therefore
![\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\\stackrel{center}{(\stackrel{x_1}{-7}~,~\stackrel{y_1}{-1})}\qquad (\stackrel{x_2}{8}~,~\stackrel{y_2}{7})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}\\\\\\\stackrel{radius}{r}=\sqrt{[8-(-7)]^2+[7-(-1)]^2}\implies r=\sqrt{(8+7)^2+(7+1)^2}\\\\\\r=\sqrt{15^2+8^2}\implies r=\sqrt{225+64}\implies \boxed{r=17}](https://tex.z-dn.net/?f=%20%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%5C%5C%5Cstackrel%7Bcenter%7D%7B%28%5Cstackrel%7Bx_1%7D%7B-7%7D~%2C~%5Cstackrel%7By_1%7D%7B-1%7D%29%7D%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B8%7D~%2C~%5Cstackrel%7By_2%7D%7B7%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%5C%5C%5C%5C%5C%5C%5Cstackrel%7Bradius%7D%7Br%7D%3D%5Csqrt%7B%5B8-%28-7%29%5D%5E2%2B%5B7-%28-1%29%5D%5E2%7D%5Cimplies%20r%3D%5Csqrt%7B%288%2B7%29%5E2%2B%287%2B1%29%5E2%7D%5C%5C%5C%5C%5C%5Cr%3D%5Csqrt%7B15%5E2%2B8%5E2%7D%5Cimplies%20%20r%3D%5Csqrt%7B225%2B64%7D%5Cimplies%20%5Cboxed%7Br%3D17%7D%20)
and since we know that x = -15 for such a point, then
![\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2\qquad center~~(\stackrel{-7}{ h},\stackrel{-1}{ k})\qquad \qquad radius=\stackrel{17}{ r}\\\\\\\[x-(-7)]^2+[y-(-1)]^2=17^2\implies (x+7)^2+(y+1)^2=289\\\\\\\stackrel{\textit{since we know x = -15}}{(-15+7)^2+(y+1)^2=289}\implies (-8)^2+\stackrel{FOIL}{(y^2+2y+1^2)}=289\\\\\\64+y^2+2y+1=289\implies y^2+2y=224\implies y^2+2y-224=0\\\\\\(y+16)(y-14)=0\implies y=\begin{cases}-16\\14\end{cases}](https://tex.z-dn.net/?f=%20%5Cbf%20%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%28x-%20h%29%5E2%2B%28y-%20k%29%5E2%3D%20r%5E2%5Cqquad%20center~~%28%5Cstackrel%7B-7%7D%7B%20h%7D%2C%5Cstackrel%7B-1%7D%7B%20k%7D%29%5Cqquad%20%5Cqquad%20radius%3D%5Cstackrel%7B17%7D%7B%20r%7D%5C%5C%5C%5C%5C%5C%5C%5Bx-%28-7%29%5D%5E2%2B%5By-%28-1%29%5D%5E2%3D17%5E2%5Cimplies%20%28x%2B7%29%5E2%2B%28y%2B1%29%5E2%3D289%5C%5C%5C%5C%5C%5C%5Cstackrel%7B%5Ctextit%7Bsince%20we%20know%20x%20%3D%20-15%7D%7D%7B%28-15%2B7%29%5E2%2B%28y%2B1%29%5E2%3D289%7D%5Cimplies%20%28-8%29%5E2%2B%5Cstackrel%7BFOIL%7D%7B%28y%5E2%2B2y%2B1%5E2%29%7D%3D289%5C%5C%5C%5C%5C%5C64%2By%5E2%2B2y%2B1%3D289%5Cimplies%20y%5E2%2B2y%3D224%5Cimplies%20y%5E2%2B2y-224%3D0%5C%5C%5C%5C%5C%5C%28y%2B16%29%28y-14%29%3D0%5Cimplies%20y%3D%5Cbegin%7Bcases%7D-16%5C%5C14%5Cend%7Bcases%7D%20)
since it's a circle, it touches x = -15 twice, check the picture below.
Answer:
0.18 sec or 3.4 sec.
For second part, it will take 3.7 sec.
First part:
Set h(t) = 17and solve for t.
-16t²+ 58t + 7= 17
-16t² + 58t - 10 = 0
Solve this quadratic equation for t. You should get 2 positive solutions. The lower value is the time to reach 17 on the way up, and the higher value is the time to reach 17 again, on the way down.
Second part:
Set h(t) = 0 and solve the resulting quadratic equation for t. You should get a negative solution (which you can discard), and a positive solution. The latter is your answer.
Answer:
x²+2x+3....
Step-by-step explanation:
The given expression is:
5x3+10x2+15x/5x
To simplify this expression take out the common of the numerator:
5x is common in all three terms of the numerator:
Like,
5x(x²+2x+3)/5x
Now the value of the common and the denominator is same. Therefore 5x will be cancel out by 5x
Then the remaining term is:
= x²+2x+3
Thus the answer is x²+2x+3....
LMFAAO it’s SSS, SAS, NEI, SAS