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3 years ago

Please help,I need it for tomorrow

Mathematics

1 answer:

Inga [223]3 years ago

7 0

Answer:

<C=26

<B=48

<D=106

<F=42

Step-by-step explanation:

I labeled all the angles to help this make more sense :)

So first we find <C, which is supplementary to <E so just do 180-154=26

Next, we know that <F and <B are equal because they form corresponding angles.

Angles <C <D and <B form a triangle. The three angles of a triangle always equal 180. Now that we know angles <C and <B, we just do 180-26-48=106.

Lastly, to find angle A, we just take 180-106-48=42.

Hope this helps

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Please help me ❤️ Me ayudan por favor

tino4ka555 [31]

Step-by-step explanation:

the given traingle is a isosoles triangle.

in a isosles traingle base angles are equal.

(8x-23)+(8x-23)+34=180(sum of <s of triangle)

or,8x-23+8x-23+34=180

or,16x-46+34=180

or,16x-12=180

or,16x=180+12

or,16x=192

or,x=192/16

Therefore, x=12

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3 years ago

[10 points] Given matrix A =  2 2 3, −6 −7 8 (a) (5 points). Show that A has no LU decomposition. (b) (5 points). Find the dec

ch4aika [34]

Answer:

Both the answers are as in the solution.

Step-by-step explanation:

As the given matrix is not in the readable form, a similar question is found online and the solution of which is attached herewith.

Part a:

Given matrix is : A = $\left[\begin{array}{ccc}0&3&4\\1&2&3\\-3&-7&8\end{array}\right]$

Here,

$det(A) =\left|\begin{array}{ccc}0&3&4\\1&2&3\\-3&-7&8\end{array}\right| = -55 \neq 0.$

Then, A is non-singular matrix.

Here, A₁₁= 0.

If we write A as LU with L lower triangular matrix and U upper triangular matrix, then A₁₁=L₁₁U₁₁.

So, As

A₁₁ = 0 gives L₁₁U₁₁= 0 ,

This indicates that either L₁₁= 0 or U₁₁ = 0.

If L₁₁= 0 or U₁₁ = 0, this would made the corresponding matrix singular, which contradicts the condition as A is non-singular.

Therefore, A has no LU decomposition.

Part b:

By the implementation of the various row operations

interchange R1 and R2

$\left[\begin{array}{ccc}1&2&3\\0&3&4\\-3&-7&8\end{array}\right]$

R3+3R1=R3

$\left[\begin{array}{ccc}1&2&3\\0&3&4\\0&-1&17\end{array}\right]$

R3+(1/3)R2 = R3

$\left[\begin{array}{ccc}1&2&3\\0&3&4\\0&0&55/3\end{array}\right]$

Therefore, U = $\left[\begin{array}{ccc}1&2&3\\0&3&4\\0&0&55/3\end{array}\right]$ .

Here, LP = E₁₂=E₃₁=-3 &E₃₂=-1/3

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

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

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

So now U is given as

$U=\left[\begin{array}{ccc}1&2&3\\0&3&4\\0&0&55/3\end{array}\right]\\L=\left[\begin{array}{ccc}1&0&0\\0&1&0\\-1/3&-3&1\end{array}\right]\\P=\left[\begin{array}{ccc}0&1&0\\1&0&0\\0&0&1\end{array}\right]\\$

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GarryVolchara [31]

Answer:

4, see below :)

Step-by-step explanation:

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3 years ago

Please help,I need it for tomorrow​

Please help,I need it for tomorrow