85% of 142 is 120.7. Rounded to 121. If 91 have already completed training, then 121-91=30
So 30 more employees must complete training.
Answer:
the statement that says each side s 32 inches long has a greater area than 48 inches long and 20 inches wide
Step-by-step explanation:
Answer:
1. LM < PN
2. AD < DC
3. m<CAB < m<CBA
4. m<1 = m<2
Step-by-step explanation:
Recall: an angle measure is relative to the length of the opposite side. That is, the longer the side opposite to an angle, the larger the measure of that angle and vice versa.
1. LM is opposite to <LNM,
PN is opposite to <NLP
m<LNM is less than m<NLP, therefore,
LM < PN
2. AD is opposite to <ABD
DC is opposite to <DBC
m<ABD is less than m<DBC, therefore,
AD < DC
3. m<CAB is opposite to CB
m<CBA is opposite to CA
CB is less than CA, therefore,
m<CAB < m<CBA
4. The side opposite to <1 is congruent to the side opposite to <2.
Therefore,
m<1 = m<2
Answer:
1. The matrix A isn't the inverse of matrix B.
2. |B|=12, |A|=12
Step-by-step explanation:
1. We want to know if matrix A is the inverse of matrix B, this means that if you do the product between B and A you have to obtain the identity matrix.
We have:
![A=\left[\begin{array}{cc}4&-2\\-1&3\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-2%5C%5C-1%263%5Cend%7Barray%7D%5Cright%5D)
and
![B=\left[\begin{array}{cc}3&2\\1&4\end{array}\right]](https://tex.z-dn.net/?f=B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%262%5C%5C1%264%5Cend%7Barray%7D%5Cright%5D)
A and B are 2×2 matrices (2 rows and 2 columns), if you multiply them you have to obtain a 2×2 matrix.
Then if A is the inverse of B:
Where,
![I=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]](https://tex.z-dn.net/?f=I%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D)
Observation:
If you have two matrices:
![A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\and\\B=\left[\begin{array}{cc}e&f\\g&h\end{array}\right]\\\\\\A.B=\left[\begin{array}{cc}(a.e+b.g)&(a.f+b.h)\\(c.e+d.g)&(c.f+d.h)\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%5C%5Cand%5C%5CB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7De%26f%5C%5Cg%26h%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CA.B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%28a.e%2Bb.g%29%26%28a.f%2Bb.h%29%5C%5C%28c.e%2Bd.g%29%26%28c.f%2Bd.h%29%5Cend%7Barray%7D%5Cright%5D)
Now:
![B.A=\left[\begin{array}{cc}3&2\\1&4\end{array}\right].\left[\begin{array}{cc}4&-2\\-1&3\end{array}\right]\\\\\\B.A=\left[\begin{array}{cc}4.3+(-2).1&4.2+(-2).4\\(-1).3+3.1&(-1).2+3.4\end{array}\right]\\\\\\B.A=\left[\begin{array}{cc}12-2&8-8\\-3+3&-2+12\end{array}\right]\\\\\\B.A=\left[\begin{array}{cc}10&0\\0&10\end{array}\right]](https://tex.z-dn.net/?f=B.A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%262%5C%5C1%264%5Cend%7Barray%7D%5Cright%5D.%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-2%5C%5C-1%263%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CB.A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4.3%2B%28-2%29.1%264.2%2B%28-2%29.4%5C%5C%28-1%29.3%2B3.1%26%28-1%29.2%2B3.4%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CB.A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D12-2%268-8%5C%5C-3%2B3%26-2%2B12%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CB.A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D10%260%5C%5C0%2610%5Cend%7Barray%7D%5Cright%5D)
![B.A=\left[\begin{array}{cc}10&0\\0&10\end{array}\right]\neq \left[\begin{array}{cc}1&0\\0&1\end{array}\right]=I\\\\\\B.A\neq I](https://tex.z-dn.net/?f=B.A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D10%260%5C%5C0%2610%5Cend%7Barray%7D%5Cright%5D%5Cneq%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%3DI%5C%5C%5C%5C%5C%5CB.A%5Cneq%20I)
Then, the matrix A isn't the inverse of matrix B.
2. If you have a matrix A:
![A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D)
The determinant of the matrix is:
Then the determinant of B is:
The determinant of A is:
Total surface area = lateral area of cone + lateral area of cylinder + area of base of cylinder:
cone: πrL = π(5cm)(13cm) = 65πcm²
cylinder: 2πrh = 2π(5cm)(12cm) = 120πcm²
base: πr² = π(5cm)² = 25πcm²
total surface area = 65πcm² + 120πcm² + 25πcm² = 210πcm²
