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Phantasy [73]
4 years ago
13

mulch to cover a rectangular garden costs 48. mulch is needed to cover a larger, similar garden. the ratio of dimensions is 2:3.

how much would it cost to cover the larger
Mathematics
2 answers:
d1i1m1o1n [39]4 years ago
5 0
\bf \qquad \qquad \textit{ratio relations}
\\\\
\begin{array}{ccccllll}
&Sides&Area&Volume\\
&-----&-----&-----\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3}
\end{array} \\\\
-----------------------------\\\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\
-------------------------------

\bf \cfrac{\stackrel{small}{garden}}{\stackrel{large}{garden}}\qquad \stackrel{sides~ratio}{\cfrac{2}{3}}\qquad \stackrel{\textit{area ratio}}{\cfrac{s^2}{s^2}}\implies \cfrac{2^2}{3^2}
\\\\\\
\cfrac{2^2}{3^2}=\cfrac{\stackrel{small~area~cost}{48}}{\stackrel{large~area~cost}{a}}\implies \cfrac{4}{9}=\cfrac{48}{a}\implies a=\cfrac{9\cdot 48}{4}

the ratio of the cost, is tandem to the ratio of the areas, because, to get the cost for the mulch, you have to first know how many square meters/feet have, and then you apply the cost evenly, now, the cost is just a factor, therefore, the ratio is retained.
Sladkaya [172]4 years ago
5 0

Answer: $108

Step-by-step explanation:

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Step-by-step explanation:

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Using Cramer's Rule, what is the value of x in the system of linear equations below?
Luden [163]

Answer:

0

Step-by-step explanation:

We find the determinant of a matrix by the method below. If we have a matrix:

\left[\begin{array}{cc}a&b\\c&d\end{array}\right]

The determinant is ad-bc

Now, using cramer's rule, we find x-value by the formula:

x=\frac{D_x}{D}

Where D is the determinant of the original problem and D_x is the determinant of the x-value matrix. How do we get those?

<u><em>To get original matrix and thus D, we set up the matrix as the coefficients of x and y (s) of both the equations and to get matrix of x-value and thus D_x, we replace the x values of the matrix with the numbers in the right hand side of the 2 equations.</em></u> We show this below:

<em />

<em>To get D:</em>

\left[\begin{array}{cc}3&4\\1&-6\end{array}\right] \\D=(3)(-6)-(1)(4)=-18-4=-22

<em>To get D_x:</em>

<em>\left[\begin{array}{cc}12&4\\-18&-6\end{array}\right] \\D_x=(12)(-6)-(-18)(4)=0</em>

<em />

<em>Putting into the formula, we get:</em>

<em>x=\frac{D_x}{D}=\frac{0}{-22}=0</em>

<em />

<em>Thus, the value of x is 0</em>

5 0
3 years ago
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