It wouldn't be able to be because since all of the the numbers that could fit is probably going to be a decimal. Let's say we find 4 consecutive numbers of 32.
Equation:
x+(x+1)+(x+2)+(x+3)=32
4x+6=32
-6 -6
4x=26
x=6.5 1. 6.5, 7.5, 8.5, 9.5
As you can see, the consecutive number is going to be a decimal.
Answer:
Triangle C
Step-by-step explanation:
Because triangle C is in the same orientation, meaning it was not flipped, it is a simple translation.
Answer: R(t) = 10 + 0.50t
Step-by-step explanation:
From the question, we've been informed that Joseph's hourly rate is $10 and that for every year he works at the grocery store, his hourly rate increases by $0.50.
Therefore, the function will be
R(t) = 10 + 0.50t
Answer:
x=75
Step-by-step explanation:
74 = x-1
+1 +1
75=x
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
