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Zolol [24]
2 years ago
13

The formula for the hypotenuse of a right triangle with legs of length a and b is shown.

Mathematics
2 answers:
fomenos2 years ago
6 0

Answer: b=\sqrt{c^{2}-a^{2}}

Step-by-step explanation:

To solve the exercise you must solve for b from the formula for the hypotenuse, as you can see below:

- Square both sides of the equation as following:

c^{2}=(\sqrt{a^{2}+b^{2}})^{2}

- Now you must subtract a² from each side of the equation, then you obtain:

c^2-a^{2}=a^{2}-a^{2}+b^{2}

c^2-a^{2}=b^{2}

- Apply square root to both sides:

\sqrt{b^{2}}=\sqrt{c^{2}-a^{2}}

Then:

b=\sqrt{c^{2}-a^{2}}

vfiekz [6]2 years ago
6 0

Answer:

see explanation

Step-by-step explanation:

Given

c = \sqrt{a^2+b^2}

Square both sides

c² = a² + b² ( subtract a² from both sides )

c² - a² = b² ( take the square root of both sides )

\sqrt{c^2-a^2} = b

⇒ b = \sqrt{c^2-a^2}

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Answer:

4 hours = 98miles

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4 0
3 years ago
Determine whether the following are subspaces of R2×2 : (a) Thesetofall2×2diagonalmatrices (b) The set of all 2 × 2 triangular m
Mrrafil [7]

A vector space V is a subspace of a vector space W if

  • V is non-empty,
  • for any two vectors v_1,v_2\inV we have v_1+v_2\in V, and
  • for any scalar k and v\in V we have kv\in V.

It's easy to show the first condition is met by all the sets in parts (a-g).

(a) is a subspace of \Bbb R^{2\times2} because adding any 2x2 diagonal matrices together, or multiplying one by some scalar, gives another diagonal matrix.

(b) and (c) are also subspaces for the same reasons.

(d) is not a subspace because \Bbb R^{2\times2} because this set of matrices does not contain the zero matrix.

(e), however, is a subspace. Any linear combination of matrices in this set always yields a matrix with 0 in row 1, column 1 entry.

(f) is a subspace. A symmetric matrix is one of the form

\begin{bmatrix}a&b\\b&c\end{bmatrix}

Adding two symmetric matrices gives another symmetric matrix:

\begin{bmatrix}a_1&b_1\\b_1&c_1\end{bmatrix}+\begin{bmatrix}a_2&b_2\\b_2&c_2\end{bmatrix}=\begin{bmatrix}a_1+a_2&b_1+b_2\\b_1+b_2&c_1+c_2\end{bmatrix}

(g) is not a subspace. Consider the matrices

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Both matrices have determinant 0, but their sum is the identity matrix with determinant 1.

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3 years ago
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Answer:

B

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Answer:

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Rate= 6.4%

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