The figure shows two intersecting chord, thus this simply can be solved using the Intersecting Chord Theorem, which states that "<span>When two chords intersect each other inside a circle, the products of their segments are equal."
Thus,
(5mxn=15mx2m)
n=6m</span>
Answer:
AD = 6.76 inches
Step-by-step explanation:
Taking the triangle ABC:
tan 45° = AB/BC
1 = AB/BC
AB = BC = 16 in
Taking the triangle DBC:
tan 30° = DB/BC
DB = BC * tan 30°
DB = 16 * tan 30°
DB = 9.24 in
Given that AD + DB = AB, then
AD = AB - DB
AD = 16 - 9.24 = 6.76 inches
Answer:
It would be a 3943.002
Step-by-step explanation:
Answer:
![B)\,\,A^{-1}=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right]](https://tex.z-dn.net/?f=B%29%5C%2C%5C%2CA%5E%7B-1%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%26-1%26-1%5C%5C-3%261%260%5C%5C-3%260%261%5Cend%7Barray%7D%5Cright%5D%20)
Step-by-step explanation:
Expanding with first row
![det(A) = \left\Bigg|\begin{array}{ccc}1&1&1\\3&4&3\\3&3&4\end{array}\right\Bigg|\\\\\\det(A)= (1)\left\Big|\begin{array}{cc}4&3\\3&4\end{array}\right\Big|-(1)\left\Big|\begin{array}{cc}3&3\\3&4\end{array}\right\Big|+(1)\left\Big|\begin{array}{cc}3&4\\3&3\end{array}\right\Big|\\\\det(A)=1[16-9]-1[12-9]+1[9-12]\\\\det(A)=7-3-3\\\\det(A)=1](https://tex.z-dn.net/?f=det%28A%29%20%3D%20%5Cleft%5CBigg%7C%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%5C%5C3%264%263%5C%5C3%263%264%5Cend%7Barray%7D%5Cright%5CBigg%7C%5C%5C%5C%5C%5C%5Cdet%28A%29%3D%20%281%29%5Cleft%5CBig%7C%5Cbegin%7Barray%7D%7Bcc%7D4%263%5C%5C3%264%5Cend%7Barray%7D%5Cright%5CBig%7C-%281%29%5Cleft%5CBig%7C%5Cbegin%7Barray%7D%7Bcc%7D3%263%5C%5C3%264%5Cend%7Barray%7D%5Cright%5CBig%7C%2B%281%29%5Cleft%5CBig%7C%5Cbegin%7Barray%7D%7Bcc%7D3%264%5C%5C3%263%5Cend%7Barray%7D%5Cright%5CBig%7C%5C%5C%5C%5Cdet%28A%29%3D1%5B16-9%5D-1%5B12-9%5D%2B1%5B9-12%5D%5C%5C%5C%5Cdet%28A%29%3D7-3-3%5C%5C%5C%5Cdet%28A%29%3D1)
To find inverse we first find cofactor matrix
Cofactor matrix is
![C=\left[\begin{array}{ccc}7&-3&3\\-1&1&0\\-1&0&1\end{array}\right] \\\\Adj(A)=C^{T}\\\\Adj(A)=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] \\\\\\A^{-1}=\frac{adj(A)}{det(A)}\\\\A^{-1}=\frac{\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] }{1}\\\\A^{-1}=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right]](https://tex.z-dn.net/?f=C%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%26-3%263%5C%5C-1%261%260%5C%5C-1%260%261%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5CAdj%28A%29%3DC%5E%7BT%7D%5C%5C%5C%5CAdj%28A%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%26-1%26-1%5C%5C-3%261%260%5C%5C-3%260%261%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5CA%5E%7B-1%7D%3D%5Cfrac%7Badj%28A%29%7D%7Bdet%28A%29%7D%5C%5C%5C%5CA%5E%7B-1%7D%3D%5Cfrac%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%26-1%26-1%5C%5C-3%261%260%5C%5C-3%260%261%5Cend%7Barray%7D%5Cright%5D%20%7D%7B1%7D%5C%5C%5C%5CA%5E%7B-1%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%26-1%26-1%5C%5C-3%261%260%5C%5C-3%260%261%5Cend%7Barray%7D%5Cright%5D)
1 is the answer I am sure of it.
