Answer:
f(x) = x - 3
Step-by-step explanation:
y = mx + b
first let's find b, the y-intercept
based on the graph, the y-intercept is (0,-3), so b = -3
immediately, we can eliminate f(x) = 3 - x and f(x) = -3x
comment any questions pls
Answer:
the answer is gay
Step-by-step explanation:
I Logan is gay cause he asked people who liked carrots and carrots are gay
<h3>
Answer: B) hyperbola</h3>
If the cutting plane was parallel to the circular base of the cones, then we would have a circular cross section (assuming the plane doesnt cut through the vertex). However, we're told than the plane intersects both nappes, or cones, so it's not possible for the plane to be parallel to the base faces. We can rule out choice A.
We can rule out choice C and choice D for similar reasons. An ellipse only forms if the plane only cuts through one cone only, which is the same story for a parabola as well.
It was better than being drawn and quartered
Answer:
Determinant are special number that can only be defined for square matrices.
Step-by-step explanation:
Determinant are particularly important for analysis. The inverse of a matrix exist, if the determinant is not equal to zero.
How to find determinant
For a 2×2 matrix
![det ( \left[\begin{array}{cc}x&y\\a&z\end{array}\right] ) = xz-ay](https://tex.z-dn.net/?f=det%20%28%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dx%26y%5C%5Ca%26z%5Cend%7Barray%7D%5Cright%5D%20%29%20%3D%20xz-ay)
For a 3×3 matrix
we first decompose it to 2×2
![det (\left[\begin{array}{ccc}k&l&m\\o&p&q\\r&s&t\end{array}\right] )\\\\= k*det(\left[\begin{array}{cc}p&q\\s&t\end{array}\right] ) - l*det(\left[\begin{array}{cc}o&q\\r&t\end{array}\right] ) + m*det(\left[\begin{array}{cc}o&p\\r&s\end{array}\right] ) \\\\=k(pt-sq) - l(ot-rq) + m(os-rp)](https://tex.z-dn.net/?f=det%20%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dk%26l%26m%5C%5Co%26p%26q%5C%5Cr%26s%26t%5Cend%7Barray%7D%5Cright%5D%20%29%5C%5C%5C%5C%3D%20k%2Adet%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dp%26q%5C%5Cs%26t%5Cend%7Barray%7D%5Cright%5D%20%29%20-%20l%2Adet%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Do%26q%5C%5Cr%26t%5Cend%7Barray%7D%5Cright%5D%20%29%20%2B%20m%2Adet%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Do%26p%5C%5Cr%26s%5Cend%7Barray%7D%5Cright%5D%20%29%20%5C%5C%5C%5C%3Dk%28pt-sq%29%20-%20l%28ot-rq%29%20%2B%20m%28os-rp%29)
Example
Find the values of λ for which the determinant is zero
![\left[\begin{array}{ccc}s&-1&0\\-1&s&-1\\0&-1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Ds%26-1%260%5C%5C-1%26s%26-1%5C%5C0%26-1%261%5Cend%7Barray%7D%5Cright%5D)
![det(\left[\begin{array}{ccc}s&-1&0\\-1&s&-1\\0&-1&1\end{array}\right])\\\\= s*det(\left[\begin{array}{cc}s&-1\\-1&1\end{array}\right] ) - (-1)*det(\left[\begin{array}{cc}-1&-1\\0&1\end{array}\right] ) + 0*det(\left[\begin{array}{cc}-1&s\\0&-1\end{array}\right] )\\\\= s(s(1)-(-1*-1)) - (-1)(-1*1 - (-1*0)) + 0\\= s(s - 1)) + 1(-1 + 0) \\=s^{2} -s-1](https://tex.z-dn.net/?f=det%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Ds%26-1%260%5C%5C-1%26s%26-1%5C%5C0%26-1%261%5Cend%7Barray%7D%5Cright%5D%29%5C%5C%5C%5C%3D%20s%2Adet%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Ds%26-1%5C%5C-1%261%5Cend%7Barray%7D%5Cright%5D%20%29%20-%20%28-1%29%2Adet%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-1%26-1%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20%29%20%2B%200%2Adet%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-1%26s%5C%5C0%26-1%5Cend%7Barray%7D%5Cright%5D%20%29%5C%5C%5C%5C%3D%20s%28s%281%29-%28-1%2A-1%29%29%20-%20%28-1%29%28-1%2A1%20-%20%28-1%2A0%29%29%20%2B%200%5C%5C%3D%20s%28s%20-%201%29%29%20%2B%201%28-1%20%2B%200%29%20%5C%5C%3Ds%5E%7B2%7D%20-s-1)
Equating the determinant to zero
s = 
* (1 ±5 )
s = 1.61 or -0.61
