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Korolek [52]
3 years ago
9

A rectangular pan has a length that is 4/3 the width. The total area of the pan is 432 in.2. What is the width of the cake pan

Mathematics
2 answers:
kirill115 [55]3 years ago
6 0

Answer:

18 inches

Step-by-step explanation:

Helga [31]3 years ago
5 0
Let width = w
length = 4w/3

Area = 432 = length*width = w*4w/3

4w^2/3 = 432
w^2 = 432*3/4 = 324
w = 18
so the width of the cake pan is 18 inches.
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Step-by-step explanation:

Let be \vec b = 2\,i+j-3\,k and \vec a = 3\,i-j, the component of \vec b parallel to \vec a is calculated by the following expression:

\vec b_{\parallel} = (\vec b \bullet \hat{a}) \cdot \hat{a}

Where \hat{a} is the unit vector of \vec a, dimensionless and \bullet is the operator of scalar product.

The unit vector of \vec a is:

\hat{a} = \frac{\vec {a}}{\|\vec a\|}

Where \|\vec {a}\| is the norm of \vec a, whose value is determined by Pythagorean Theorem.

The component of \vec{b} parallel to \vec {a} is:

\|\vec {a}\| = \sqrt{3^{2}+(-1)^{2}+0^{2}}

\|\vec {a}\| = \sqrt{10}

\hat{a} = \frac{1}{\sqrt{10}} \cdot (3\,i-j)

\hat{a} = \frac{3}{\sqrt{10}}\,i -\frac{1}{\sqrt{10}} \,j

\vec{b}\bullet \hat{a} = (2)\cdot \left(\frac{3}{\sqrt{10}} \right)+(1)\cdot \left(-\frac{1}{\sqrt{10}} \right)+(-3)\cdot \left(0\right)

\vec b \bullet \hat{a} = \frac{5}{\sqrt{10}}

\vec b_{\parallel} = \frac{5}{\sqrt{10}}\cdot \left(\frac{3}{\sqrt{10}}\,i-\frac{1}{\sqrt{10}}\,j  \right)

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\vec b _{\perp} = \frac{1}{2}\,i+\frac{3}{2}\,j-3\,k

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