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insens350 [35]
3 years ago
7

Requiem needs to know if the triangle shown is a right triangle. Which equation could he use to help ?

Mathematics
1 answer:
finlep [7]3 years ago
3 0

Answer:

Pythagoras theorem: hypotenuse² = opposite² + adjacent²

Step-by-step explanation:

To know if a triangle is a right angle, we need to have the length of each sides of the triangle. Then we would apply Pythagoras theorem to determine if it is truly a right angled triangle.

Pythagoras theorem is a theorem in the form of an equation which shows the relationship between the sides of a right angled triangle.

let the sides of the right angled triangle be:

opposite =a, adjacent = b and hypotenuse = c

Using Pythagoras theorem

hypotenuse² = opposite² + adjacent²

c² = a² + b²

If the left hand side of the equation = the right hand side of the equation, it is a right angled triangle.

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The vectors in S form a basis of P_2 if they are mutually linearly independent and span P_2.

To check for independence, we can compute the Wronskian determinant:

\begin{vmatrix}x^2+3x+1&2x^2+x-1&4\\2x+3&4x+1&0\\2&4&0\end{vmatrix}=4\begin{vmatrix}2x+3&4x+1\\2&4\end{vmatrix}=40\neq0

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To check if they span P_2, you need to show that any vector in P_2 can be expressed as a linear combination of the vectors in S. We can write an arbitrary vector in P_2 as

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Then we need to show that there is always some choice of scalars k_1,k_2,k_3 such that

k_1(x^2+3x+1)+k_2(2x^2+x-1)+k_34=p

This is equivalent to solving

(k_1+2k_2)x^2+(3k_1+k_2)x+(k_1-k_2+4k_3)=ax^2+bx+c

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\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}

This has a solution if the coefficient matrix on the left is invertible. It is, because

\begin{vmatrix}1&1&0\\3&1&0\\1&-1&4\end{vmatrix}=4\begin{vmatrix}1&2\\3&1\end{vmatrix}=-20\neq0

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Compute the inverse any way you like; you should get

\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}=-\dfrac1{20}\begin{bmatrix}4&-8&0\\-12&4&0\\-4&3&-5\end{bmatrix}

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A solution exists for any choice of a,b,c, so the vectors in S indeed span P_2.

The vectors in S are independent and span P_2, so S forms a basis of P_2.

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