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geniusboy [140]
1 year ago
11

3. Find how many numbers between 232 and 252.​

Mathematics
1 answer:
igomit [66]1 year ago
7 0

Answer:

252-232=20-1=19

Step-by-step explanation:

the question said in between so you don't count the first and the last numbers

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1 year ago
Complete the proof by providing the missing statement and reasons
Lisa [10]

Answer:

In triangle SHD and triangle STD.

\overline{SD} \perp \overline{HT}          [Side]

Since, a line is said to be perpendicular to another line if the two lines intersect at a right angle.

⇒ \angle SDH = \angle SDT = 90^{\circ}

\overline{SH} \cong \overline{ST}       [leg]               [Given]

Reflexive property states that the value is equal to itself.

\overline{SD} \cong \overline{SD}       [Leg]       [Reflexive property]

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6 0
3 years ago
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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Answer:

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Step-by-step explanation:

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3 years ago
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2 years ago
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