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2 years ago

9

12(4x−5)+5/2=10 solve for x pls help!!!!

1 answer:

True [87]2 years ago

5 0

Answer:

Step-by-step explanation:

$12(4x - 5) + \frac{5}{2}=10$

Multiply the whole equation by 2

$2*12(4x -5) + \frac{5}{2}*2=10*2\\\\24(4x-5) + 2 = 20\\$

24*4x - 5*24 + 2 = 20

96x - 120 + 2 = 20

96x - 118 = 20

Add 118 to both sides

96x = 20 +118

96x = 138

Divide both sides by 96

x = 138/96 = 23/16

$x = 1 \frac{7}{16}$

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An arched drainage culvert can be modelled by the function:

Zanzabum

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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