Answer: parallel because the slopes are the same
Step-by-step explanation:
the formula is f=(c*20)+30 so 14 degrees farhenhight
Answer: the equation is
5x^2 -30x - 25
Step-by-step explanation:
A quadratic equation is one in which the highest power of the unknown is 2.
The general form of a quadratic equation is expressed as
ax^2 + bx + c
Where c is a constant and a is the leading coefficient
Assuming we want to write the quadratic equation in x, from the information given, the given roots are 5 and 1 and the leading coefficient is 5. We will just multiply the expression by the leading coefficient.
Therefore, the linear factors of the quadratic will be (x-5) and (x-1)
With the leading coefficient as 5, the equation becomes
5(x-5)(x-1)
= 5(x^2 - x - 5x + 5)
= 5(x^2 - 6x + 5)
= 5x^2 -30x - 25
Let the two integers be x and x+1
(x+1)^2-x^2=21
x^2+2x+1-x^2=21
2x+1=21
2x=20
x=10
then x=10, x+1
10+1=11
the two integers are 10, 11
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.
