Suppose DK is an angle bisector of △DEF. Given: EK=2, FK=5, DF=10. Find DE.

Mathematics

1 answer:

Inessa [10]3 years ago

8 0

Answer:

DE = 4

Step-by-step explanation:

The bisector divides the triangle into proportional segments, so ...

FK/DF = EK/DE

5/10 = 2/DE . . . . substitute given values

DE = 4 . . . . . multiply by 2DE

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Let A = −2 2 1 −3 1 1 2 0 −1 and B = 2 −1 0 1 2 1 −1 −2 4 . Use the matrix-column representation of the product to write each co

BigorU [14]

Answer:

AB = $\left[\begin{array}{ccc}-3&4&6\\-6&3&5\\5&0&-4\end{array}\right]$

Each column of AB is written as a linear combination of columns of Matrix A in the explanation below.

Step-by-step explanation:

A = $\left[\begin{array}{ccc}-2&2&1\\-3&1&1\\2&0&-1\end{array}\right]$

B= $\left[\begin{array}{ccc}2&-1&0\\1&2&1\\-1&-2&4\end{array}\right]$

We need to write each column of AB as a linear combination of the columns of A so we will multiply each column of A with each column element of B to get the column of AB. So,

AB Column 1 = 2 * $\left[\begin{array}{ccc}-2\\-3\\2\end{array}\right]$ + 1 $\left[\begin{array}{ccc}2\\1\\0\end{array}\right]$ + (-1) $\left[\begin{array}{ccc}1\\1\\-1\end{array}\right]$ = $\left[\begin{array}{ccc}-3\\-6\\5\end{array}\right]$

AB Column 2 = (-1) $\left[\begin{array}{ccc}-2\\-3\\2\end{array}\right]$ + 2 $\left[\begin{array}{ccc}2\\1\\0\end{array}\right]$ + (-2) $\left[\begin{array}{ccc}1\\1\\-1\end{array}\right]$ = $\left[\begin{array}{ccc}4\\3\\0\end{array}\right]$

AB Column 3 = (0) $\left[\begin{array}{ccc}-2\\-3\\2\end{array}\right]$ + (1) $\left[\begin{array}{ccc}2\\1\\0\end{array}\right]$ + 4 $\left[\begin{array}{ccc}1\\1\\-1\end{array}\right]$ = $\left[\begin{array}{ccc}6\\5\\-4\end{array}\right]$