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3 years ago

11

Find the distance between (-1,-2) and (8,9)

2 answers:

Rom4ik [11]3 years ago

7 0

Answer:

$\sqrt{202}$ units or 14.2 units (rounded to the nearest tenth)

Step-by-step explanation:

To find the distance between a pair of two points, use the distance formula $d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ . Substitute the x and y values of (-1,-2) and (8,9) into the formula and simplify:

$d = \sqrt{(-1-8)^2+(-2-9)^2} \\d = \sqrt{(-9)^2+(-11)^2} \\d = \sqrt{81+121} \\d= \sqrt{202}$

So, as an exact answer, the distance is $\sqrt{202}$ units.

(To find the decimal approximate of that answer, enter it into a calculator. This would make the decimal approximate 14.2 units, when rounded to the nearest tenth.)

serious [3.7K]3 years ago

6 0

Answer:

14.2

Step-by-step explanation:

All you need to do is plug the coordinates into the distance formula.

√(x2 - x1)² + (y2 - y1)²

√[8 - (-1)]² + [9 - (-2)]²

√(9)² + (11)²

√81 + 121

√202

= 14.2

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2 years ago

An arched drainage culvert can be modelled by the function:

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Substitute 136.5 for y.

$136.5=\frac{x(191-x)}{60}$

$8190=x(191-x)$

$8190=191x-x^{2}$

$x^{2} -191x+8190=0$

Compare this equation with $ax^{2} +bx+c=0$

a = 1, b = - 191, c = 8190

$x=\frac{-b+\sqrt{b^{2}-4ac } }{2}$ or

$x=\frac{-b-\sqrt{b^{2}-4ac } }{2}$

$x=\frac{191+\sqrt{191^{2}-4(1)(8190) } }{2}$ or

$x=\frac{191-\sqrt{191^{2}-4(1)(8190) } }{2}$

$x=\frac{191+\sqrt{36481-32760 } }{2}$ or

$x=\frac{191-\sqrt{36481-32760 } }{2}$

$x=\frac{191+\sqrt{3721 } }{2}$ or

$x=\frac{191-\sqrt{3721 } }{2}$

$x=\frac{191+61}{2}$ or

$x=\frac{191-61}{2}$

$x=\frac{252}{2}$ or

$x=\frac{130}{2}$

x = 126 or x = 65

Hence, the points on the curve are (65, 136.5) and (126, 136.5).

The width of the air space is the distance between these points.

Width = $\sqrt{(x_{2}- x_{1})^{2} +( y_{2}- y_{1}) ^{2}$

= $\sqrt{(126- 65)^{2} +(136.5-136.5) ^{2}$

= 126 - 65

= 61

Hence, width of the air space is 61m.

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2 years ago