Solve using the pathagorean theorem

Mathematics

1 answer:

seropon [69]2 years ago

4 0

Answer:

2 sqrt(13)

Step-by-step explanation:

We can find the hypotenuse using the Pythagorean theorem

a^2 + b^2 = c^2 where a and b are the legs and c is the hypotenuse

4^2 + 6^2 = c^2

16+36 = c^2

52 = c^2

Taking the square root of each side

sqrt (52) = c

sqrt(4*13) = c

2 sqrt(13) = c

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Seperate the number 53 into 3 parts so that the second number is 3 less than the first number and the third number is twice the

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Hello,

Let a the first, b the second and c the third.
a+b+c=53
b=a/3
c=2a
==>a+a/3+2a=53
==>10a/3=53
==>a=15.9
c=2*a=2*15.9=31.8
b=a/3=15.9/3=5.3

Proof: a+b+c=15.9+5.3+31.8=53

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ANSWER

EXPLANATION

For a matrix A of order n×n, the cofactor $C_{ij}$ of element $a_{ij}$ is defined to be

$C_{ij} = (-1)^{i+j} M_{ij}$

$M_{ij}$ is the minor of element $a_{ij}$ equal to the determinant of the matrix we get by taking matrix A and deleting row i and column j.

Here, we have

$C_{11} = (-1)^{1+1} M_{11} = M_{11}$

M₁₁ is the determinant of the matrix that is matrix A with row 1 and column 1 removed. The bold entries are the row and the column we delete.

$\begin{aligned} A=\begin{bmatrix} \bf 1 & \bf -6 & \bf -4\\ \bf 7 & 0 & -3 \\ \bf -9 & 8 & -8 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 0&-3 \\ 8&-8 \end{bmatrix} \right) \end{aligned}$