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nekit [7.7K]
3 years ago
11

Can you solve #7 if ur not in 8th grade or higher you probably won’t understand but if you do your smart:)

Mathematics
1 answer:
BlackZzzverrR [31]3 years ago
7 0
The other leg measures 20.5 m.

Since this is a right triangle, we can use the Pythagorean theorem to solve this.  The Pythagorean theorem says that the sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse, or 

a²+b²=c², where a and b are the legs and c is the hypotenuse.

Plugging in the value of the one known leg and the hypotenuse, we have
32²+b²=38²
32*32 + b² = 38*38
1024 + b² = 1444

Subtract 1024 from both sides:
1024 + b² - 1024 = 1444 - 1024
b² = 420

Take the square root of both sides:
√b² = √420
b ≈ 20.5
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[(a)^2-(4^2)](a+4)

Step-by-step explanation:

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Read 2 more answers
skew-symmetric 3 x 3 matrices form as subspace of all 3 x 3 matrices and find a basis for this subspace.
Neporo4naja [7]

Answer:

a) ∝A ∈ W

so by subspace, W is subspace of 3 × 3 matrix

b) therefore Basis of W is

={ {\left[\begin{array}{ccc}0&1&0\\-1&0&0\\0&0&0\end{array}\right] ,\left[\begin{array}{ccc}0&0&1\\0&0&0\\-1&0&0\end{array}\right] ,\left[\begin{array}{ccc}0&0&0\\0&0&1\\0&-1&0\end{array}\right]}

Step-by-step explanation:

Given the data in the question;

W = { A| Air Skew symmetric matrix}

= {A | A = -A^T }

A ; O⁻ = -O⁻^T        O⁻ : Zero mstrix

O⁻ ∈ W

now let A, B ∈ W

A = -A^T       B = -B^T

(A+B)^T = A^T + B^T

= -A - B

- ( A + B )

⇒ A + B = -( A + B)^T

∴ A + B ∈ W.

∝ ∈ | R

(∝.A)^T = ∝A^T

= ∝( -A)

= -( ∝A)

(∝A) = -( ∝A)^T

∴ ∝A ∈ W

so by subspace, W is subspace of 3 × 3 matrix

A ∈ W

A = -AT

A = \left[\begin{array}{ccc}o&a&b\\-a&o&c\\-b&-c&0\end{array}\right]

= a\left[\begin{array}{ccc}0&1&0\\-1&0&0\\0&0&0\end{array}\right] +b\left[\begin{array}{ccc}0&0&1\\0&0&0\\-1&0&0\end{array}\right] +c\left[\begin{array}{ccc}0&0&0\\0&0&1\\0&-1&0\end{array}\right]

therefore Basis of W is

={ {\left[\begin{array}{ccc}0&1&0\\-1&0&0\\0&0&0\end{array}\right] ,\left[\begin{array}{ccc}0&0&1\\0&0&0\\-1&0&0\end{array}\right] ,\left[\begin{array}{ccc}0&0&0\\0&0&1\\0&-1&0\end{array}\right]}

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