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GaryK [48]
3 years ago
14

Please Help

Mathematics
1 answer:
Fudgin [204]3 years ago
3 0

subtract 1/4 from 7/8

find common denominator, in this case 8

1/4 = 2/8

now you have 7/8 - 2/8 = 5/8 hour left

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Step-by-step explanation:

<u>Step 1:  Convert improper fraction into a mixed number</u>

\frac{7}{6}

\frac{6}{6} + \frac{1}{6}

1 + \frac{1}{6}

1\frac{1}{6}

Answer: 1\frac{1}{6}

6 0
3 years ago
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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.



5 0
3 years ago
Can someone please solve this​
Svetradugi [14.3K]

Answer:

The answer is the 3rd one.

Step-by-step explanation:

3 times 3 is 9

4 times 4 is 16

13 is in between those 2 numbers

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2 years ago
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