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3 years ago

Find the perimeter of the triangle with coordinates A(5,2), B(5,4), and C(1,1). Round to the nearest tenth.

Mathematics

1 answer:

faltersainse [42]3 years ago

4 0

Answer:

11.1 units

Step-by-step explanation:

We solve for this using the formula when using coordinates (x1 , y1) and (x2, y2)

= √(x2 - x1)² + (y2 - y1)²

A(5,2), B(5,4), and C(1,1).

For AB = √(x2 - x1)² + (y2 - y1)²

= A(5,2), B(5,4)

= √(5 - 5)² +(4 - 2)²

= √ 0² + 2²

= √4

= 2 units

For BC = √(x2 - x1)² + (y2 - y1)²

= B(5,4), C(1,1)

= √(1 - 5)² +(1 - 4)²

= √ -4² + -3²

= √16 + 9

= √25

= 5 units

For AC = √(x2 - x1)² + (y2 - y1)²

A(5,2), C(1,1)

= √(1 - 5)² + (1 - 2)²

= √-4² + -1²

= √16 + 1

= √17

= 4.1231056256 units

The Formula for the Perimeter of Triangle = Side AB + Side BC + Side AC

= 2 units + 5 units + 4.1231056256 units

= 11.1231056256 units.

Approximately the Perimeter of a Triangle to the nearest tenth = 11.1units

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Let V be the volume of the solid obtained by rotating about the y-axis the region bounded y = sqrt(25x) and y = x^2/25. Find V b

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Explanation:

Let $f(x) = \sqrt{25x}$ and $g(x) = \frac{x^2}{25}$ . The differential volume dV of the cylindrical shells is given by

$dV = 2\pi x[f(x) - g(x)]dx$

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$\displaystyle V = 2\pi\int{x[f(x) - g(x)]}dx$

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$\sqrt{25x} = \dfrac{x^2}{25}$

We can clearly see that x = 0 is one of the solutions. For the other solution/limit, let's solve for x by first taking the square of the equation above:

$25x = \dfrac{x^4}{(25)^2} \Rightarrow \dfrac{x^3}{(25)^3} = 1$

$x^3 =(25)^3 \Rightarrow x = \pm25$

Since we are rotating the functions around the y-axis, we are going to use the x = 25 solution as one of the limits. So the expression for the volume of revolution around the y-axis is

$\displaystyle V = 2\pi\int_0^{25}{x\left(\sqrt{25x} - \frac{x^2}{25}\right)}dx$