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AveGali [126]

Answer:

FD≈25.94.. rounded = 26

Step-by-step explanation:

FD²=12²+(4x+11)²

FD²=144+16x²+88x+121

FD²=265+16x²+88x

also

FD²=12²+(13x-16)²

FD²=144+169x²-416x+256

FD²=400+169x²-416x

thus

265+16x²+88x = 400+169x²-416x

16x²-169x²+88x+416x+265-400 = 0

-153x²+504x-135 = 0

we will solve this quadratic equation by suing the quadratic formula to find x

x=(-504±sqrt(504²-4(-153)(-135)))/2(-153)

x=(-504± $\sqrt{504^{2}-4(-153)(-135) }$ )/2(-153)

x=(-504± $\sqrt{254016-82620 }$ )/-306

x=(-504± $\sqrt{171396}$ )/-306

x=(-504±414)/-306

x=(-504+414)/-306 and x=(-504-414)/-306

x=-90/-306 and x=-918/-306

x= 5/17 , 3

substituting x by the roots we found

check for 5/17:

4x+11 = 4×(5/17)+11 = (20/17)+11 = (20+187)/17 = 207/17 ≈ 12.17..

13x-16 = 13×(5/17)-16 = (65/17)-16 = (65-272)/17 = -207/17 ≈ -12.17..

check for 3:

4x+11 = 4×3+11 = 12+11 = 23

13x-16 = 13×3-16 = 23

thus 3 is the right root

therfore

ED=23 and CD=23

FD²=FE²+ED² or FD²=FC²+CD²

FD²=12²+23²

FD²=144+529

FD²=673

FD=√673

FD≈25.94.. rounded = 26

5 0

2 years ago

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Find the inverse of the given matrix, if it exists.Aequals=left bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 0 3

BabaBlast [244]

Answer:

$A^{-1}=\left[ \begin{array}{ccc} \frac{1}{9} & \frac{4}{27} & - \frac{2}{27} \\\\ \frac{8}{9} & \frac{5}{27} & \frac{11}{27} \\\\ - \frac{4}{9} & \frac{2}{27} & - \frac{1}{27} \end{array} \right]$

Step-by-step explanation:

We want to find the inverse of $A=\left[ \begin{array}{ccc} 1 & 0 & -2 \\\\ 4 & 1 & 3 \\\\ -4 & 2 & 3 \end{array} \right]$

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 4&1&3&0&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]$

Subtract row 1 multiplied by 4 from row 2

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]$

Add row 1 multiplied by 4 to row 3

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&2&-5&4&0&1\end{array}\right]$

Subtract row 2 multiplied by 2 from row 3

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&-27&12&-2&1\end{array}\right]$

Divide row 3 by −27

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

Add row 3 multiplied by 2 to row 1

$\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

Subtract row 3 multiplied by 11 from row 2

$\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&0&\frac{8}{9}&\frac{5}{27}&\frac{11}{27} \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$