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2 years ago

1. Find the perimeter and area of a rectangle with a length of 3.2 m and a width of 1.5 m.

1 answer:

5 0

<span>1. D) p=9.4 m; A= 4.8 m^2 2. C) 189 ft^2 1. Find the perimeter and area of a rectangle with a length of 3.2 m and a width of 1.5 m. The perimeter of a rectangle is simply twice the sum of it's length and width, so 2(3.2m + 1.5 m) = 2*4.7 = 9.4 m The area of a rectangle is the product of its length and width, so 3.2 m * 1.5 m = 4.8 m^2 Of the available choices, D matches with p=9.4 m; A= 4.8 m^2 2. Find the area of a triangle with a base of 27 feet and a height of 14 feet. The area of a triangle is 1/2 base times height, so 0.5 * 27 ft * 14 ft = 189 ft^2 Of the available choices, C matches with 189 ft^2</span>

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Step-by-step explanation:

Given:

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Required:

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Open the bracket

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Integrate with respect to y

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Substitute 0 and 3 for y

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$c\int\limits^3_0 {(\frac{9x}{2} +\frac{27x}{3} ) \, dx = 1$

Add fraction

$c\int\limits^3_0 {(\frac{27x + 54x}{6}) \, dx = 1$

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Rewrite;

$c\int\limits^3_0 (81x * \frac{1}{6}) \, dx = 1$

The $\frac{1}{6}$ is a constant, so it can be removed from the integral sign to give

$c * \frac{1}{6}\int\limits^3_0 (81x ) \, dx = 1$

$\frac{c}{6}\int\limits^3_0 (81x ) \, dx = 1$

Integrate with respect to x

$\frac{c}{6} * \frac{81x^{2}}{2} (0,3) = 1$

Substitute 0 and 3 for x

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$\frac{c}{6} * \frac{81 * 9 - 0}{2} = 1$

$\frac{c}{6} * \frac{729}{2} = 1$

$\frac{729c}{12} = 1$

Multiply both sides by $\frac{12}{729}$

$c = \frac{12}{729}$

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3 years ago

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Answer:

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Step-by-step explanation:

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===================================================

Given:

$(-xy)^3 (xz) =ax^by^cz^d$

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Comparing the right hand side with the left hand side

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∴ c = 3

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2 years ago