Answer:
x=3
Step-by-step explanation:
3 squared is 9. 9 plus 2 times 2=22 each of the sides on the angle are 11. 22 plus 22= 44. then 3 -1 equals 2. 44 minus 4 equals 40. equaling the area
0.088
Step-by-step explanation:
The total number of balls in the urn is;
8 + 5 = 12
Because the pink balls are 8, then the probability of picking a pink ball from the urn is;
8/12
To get the probability that all 6 balls drawn from the urn are pink, we will use the AND probability rule of the mutually exclusive events which means we’ll multiply the probabilities of each of the six pink balls;
8/12 * 8/12 * 8/12 * 8/12 * 8/12 * 8/12
= 0.088
Learn More:
For more on probabilities check out;
brainly.com/question/11248705
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Answer:
8/35 cubic inches
Step-by-step explanation:
Volume is length x width x height. To find the answer, we must multiply all values together. 2/3 times 3/5 ix 6/15. Let's simplify this before multiplying further. 6/15 can also be written as 2/5. 2/5 times 4/7 is 8/35 which cannot be simplified. 8/35 is the final answer.
I might be wrong but im pretty sure the answer is B.
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
