What 6.72 divided by 8

2 answers:

6 0

When dividing a number with a decimal remember to bring the decimal up. Put the first number under the house,them put the second number outside the house. Bring the decimal straight up and then divided like normal.

swat323 years ago

3 0

The answer will be 0.84

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Raj encoded a secret phrase using matrix multiplication. Using A = 1, B = 2, C = 3, and so on, he multiplied the clear text code

Natali5045456 [20]

Answer:

THE CORN IS YUMMY

Step-by-step explanation:

Understanding the question:

Raj encoded a secret phrase using matrix multiplication. For that he first converted each letter in that phrase to integers using the following codes:

A=1,B=2,C=3,D=4,E=5,F=6,G=7,H=8,I=9,J=10,K=11,L=12,M=13,N=14,O=15,P=16,Q=17,R=18,S=19,T=20,U=21,V=22,W=23,X=24,Y=25,Z=26

We suppose that Raj had a matrix X formed by letters of that phrase encoded in integers/numbers. And he multiplied that matrix X with matrix C to get an encoded matrix, let us say 'encoded X'.

Where; C = $\left[\begin{array}{ccc}2&5\\1&2\end{array}\right]$

and encoded X = $\left[\begin{array}{ccc}85&111&135&111&95&101&153\\38&48&55&45&41&44&64\end{array}\right]$

What we need now is X.

Raj got from the phrase to encoded matrix by:

1. Encoding each letter into numbers and forming a matrix X.

2. Multiplying that matrix with another Matrix C to obtain the given encoded matrix.

To find the secret phrase we will reverse the steps as:

1. Multiplying the encoded matrix by 'Inverse of Matrix C' to get matrix X.

2. Converting the elements of X into alphabets.

Inverse of C = (Adjoint of C)/Determinant(C)

Adjoint of $\left[\begin{array}{ccc}2&5\\1&2\end{array}\right]$ = $\left[\begin{array}{ccc}2&-5\\-1&2\end{array}\right]$

Determinant of C = |C| = ad - bc = 2*2 - 5*1 = -1

Inverse of C = $\left[\begin{array}{ccc}2&-5\\-1&2\end{array}\right]$ / (-1)

= $\left[\begin{array}{ccc}-2&5\\1&-2\end{array}\right]$

Multiplying Inverse(C) and encoded X = $\left[\begin{array}{ccc}-2&5\\1&-2\end{array}\right]$ $\left[\begin{array}{ccc}85&111&135&111&95&101&153\\38&48&55&45&41&44&64\end{array}\right]$

= $\left[\begin{array}{ccc}20&8&5&3&15&18&14\\9&19&25&21&13&13&25\end{array}\right]$