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marishachu [46]
3 years ago
11

The conversion of square feet to square yards can be represented by direct variation. Three square yards are

Mathematics
1 answer:
grandymaker [24]3 years ago
4 0
<h3>The constant of variation is \frac{1}{9}</h3><h3>The equation is y=\frac{1}{9}x</h3>

<em><u>Solution:</u></em>

Given that,

The conversion of square feet to square yards can be represented by direct variation

Let y represent the number of square yards

Let x be the number of square feet

y \propto x

y = kx ------ eqn 1

Where, "k" is the constant of variation

Three square yards are  equivalent to 27 square feet

Substitute, y = 3 and x = 27 in eqn 1

3 = k \times 27\\\\k = \frac{3}{27}\\\\k = \frac{1}{9}

Thus the constant of variation is \frac{1}{9}

<em><u>What is the equation representing the direct variation?</u></em>

Substitute k = 1/9 in eqn 1

y = \frac{1}{9}x

Thus the equation is found

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The vector (a) is a multiple of the vector (2i +3j) and (b) is a multiple of (2i+5j) The sum (a+b) is a multiple of the vector (
kow [346]

Answer:

\|a\| = 5\sqrt{13}.

\|b\| = 3\sqrt{29}.

Step-by-step explanation:

Let m,n, and k be scalars such that:

\displaystyle a = m\, (2\, \vec{i} + 3\, \vec{j}) = m\, \begin{bmatrix}2 \\ 3\end{bmatrix}.

\displaystyle b = n\, (2\, \vec{i} + 5\, \vec{j}) = n\, \begin{bmatrix}2 \\ 5\end{bmatrix}.

\displaystyle (a + b) = k\, (8\, \vec{i} + 15\, \vec{j}) = k\, \begin{bmatrix}8 \\ 15\end{bmatrix}.

The question states that \| a + b \| = 34. In other words:

k\, \sqrt{8^{2} + 15^{2}} = 34.

k^{2} \, (8^{2} + 15^{2}) = 34^{2}.

289\, k^{2} = 34^{2}.

Make use of the fact that 289 = 17^{2} whereas 34 = 2 \times 17.

\begin{aligned}17^{2}\, k^{2} &= 34^{2}\\ &= (2 \times 17)^{2} \\ &= 2^{2} \times 17^{2} \end{aligned}.

k^{2} = 2^{2}.

The question also states that the scalar multiple here is positive. Hence, k = 2.

Therefore:

\begin{aligned} (a + b) &= k\, (8\, \vec{i} + 15\, \vec{j}) \\ &= 2\, (8\, \vec{i} + 15\, \vec{j}) \\ &= 16\, \vec{i} + 30\, \vec{j}\\ &= \begin{bmatrix}16 \\ 30 \end{bmatrix}\end{aligned}.

(a + b) could also be expressed in terms of m and n:

\begin{aligned} a + b &= m\, (2\, \vec{i} + 3\, \vec{j}) + n\, (2\, \vec{i} + 5\, \vec{j}) \\ &= (2\, m + 2\, n) \, \vec{i} + (3\, m + 5\, n) \, \vec{j} \end{aligned}.

\begin{aligned} a + b &= m\, \begin{bmatrix}2\\ 3 \end{bmatrix} + n\, \begin{bmatrix} 2\\ 5 \end{bmatrix} \\ &= \begin{bmatrix}2\, m + 2\, n \\ 3\, m + 5\, n\end{bmatrix}\end{aligned}.

Equate the two expressions and solve for m and n:

\begin{cases}2\, m + 2\, n = 16 \\ 3\, m + 5\, n = 30\end{cases}.

\begin{cases}m = 5 \\ n = 3\end{cases}.

Hence:

\begin{aligned} \| a \| &= \| m\, (2\, \vec{i} + 3\, \vec{j})\| \\ &= m\, \| (2\, \vec{i} + 3\, \vec{j}) \| \\ &= 5\, \sqrt{2^{2} + 3^{2}} = 5 \sqrt{13}\end{aligned}.

\begin{aligned} \| b \| &= \| n\, (2\, \vec{i} + 5\, \vec{j})\| \\ &= n\, \| (2\, \vec{i} + 5\, \vec{j}) \| \\ &= 3\, \sqrt{2^{2} + 5^{2}} = 3 \sqrt{29}\end{aligned}.

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