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LuckyWell [14K]

3 years ago

6

The heat in the house is set to keep the minimum and maximum temperatures (in degrees Fahrenheit) according to the equation |x –

72.5| = 4. What are the minimum and maximum temperatures in the house? 72.5°F and 76.5°F 68.5°F and 76.5°F 70.5°F and 74.5°F 72.5°F and 74.5°F

1 answer:

dmitriy555 [2]3 years ago

5 0

$|x-72.5|=4 \\
x-72.5=-4 \ \lor \ x-72.5=4 \\
x=-4+72.5 \ \lor \ x=4+72.5 \\
x=68.5 \ \lor \ x=76.5$

The minimum and maximum temperatures in the house are 68.5°F and 76.5°.

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What is/are the classification of functions that has its inverse?

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Answer:

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3 years ago

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<u>Step-by-step explanation:</u>

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3 years ago

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

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

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

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

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Hence, width of the air space is 61m.

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3 years ago

What's LCD<br> 5/6<br> 3/18<br> 7/9<br> 1/126

Rudiy27

Answer:

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Step-by-step explanation:

One way to do this is to see if 126 is divisible by 6,9 and 18. If so then the LCD is 126.

Use your calculator to check.

It is 126.

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3 years ago

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3 years ago