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2 years ago

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The temperature is 60°F. The temperature will decrease by 3°F each hours. Let h be the number of hours. When will the temperatur

e be below 32° F? Write an inequality for this problem.

1 answer:

lisabon 2012 [21]2 years ago

7 0

Answer:

For the temperature to be below 32 F, following inequality must be met,

$h\geq 9.333$

Step-by-step explanation:

Let "t" be the temperature at any time.

General format of a linear equation can be given as,

y= mx + c

Where,

y ⇒ dependent variable ( in this case temperature t)

m⇒ Gradient ( the rate at which the temperature drops)

x⇒ independent variable ( hours )

c⇒ intercept ( this is the starting temperature)

Now we can substitute the given values to the above format to arrive at the equation,

$t=-3h+60$

We can re-arrange the equations as ,

$3h=60-t$

= $h=20-\frac{t}{3}$

We can write this as an inequality when t = 32 F

$h\geq 20-\frac{32}{3}$

= $h\geq \frac{28}{3}$

= $h\geq 9.333$

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Harlamova29_29 [7]

Answer:

$\huge{\purple {r= 2\times\sqrt[3]3}}$

$\huge 2\times \sqrt [3]3 = 2.88$

Step-by-step explanation:

For solid iron sphere:
radius (r) = 2 cm (Given)

Formula for $V_{sphere}$ is given as:

$V_{sphere} =\frac{4}{3}\pi r^3$

$\implies V_{sphere} =\frac{4}{3}\pi (2)^3$

$\implies V_{sphere} =\frac{32}{3}\pi \:cm^3$

For cone:
r : h = 3 : 4 (Given)
Let r = 3x & h = 4x

Formula for $V_{cone}$ is given as:

$V_{cone} =\frac{1}{3}\pi r^2h$

$\implies V_{cone} =\frac{1}{3}\pi (3x)^2(4x)$

$\implies V_{cone} =\frac{1}{3}\pi (36x^3)$

$\implies V_{cone} =12\pi x^3\: cm^3$

It is given that: iron sphere is melted and recasted in a solid right circular cone of same volume
$\implies V_{cone} = V_{sphere}$

$\implies 12\cancel{\pi} x^3= \frac{32}{3}\cancel{\pi}$

$\implies 12x^3= \frac{32}{3}$

$\implies x^3= \frac{32}{36}$

$\implies x^3= \frac{8}{9}$

$\implies x= \sqrt[3]{\frac{8}{3^2}}$

$\implies x={\frac{2}{ \sqrt[3]{3^2}}}$

$\because r = 3x$

$\implies r=3\times {\frac{2}{ \sqrt[3]{3^2}}}$

$\implies r=3\times 2(3)^{-\frac{2}{3}}$

$\implies r= 2\times (3)^{1-\frac{2}{3}}$

$\implies r= 2\times (3)^{\frac{1}{3}}$

$\implies \huge{\purple {r= 2\times\sqrt[3]3}}$
Assuming log on both sides, we find:

$log r = log (2\times \sqrt [3]3)$

$log r = log (2\times 3^{\frac{1}{3}})$

$log r = log 2+ log 3^{\frac{1}{3}}$

$log r = log 2+ \frac{1}{3}log 3$

$log r = 0.4600704139$

Taking antilog on both sides, we find:

$antilog(log r )= antilog(0.4600704139)$

$\implies r = 2.8844991406$

$\implies \huge \red{r = 2.88\: cm}$

$\implies 2\times \sqrt [3]3 = 2.88$

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Hope this helped

Hope this helped ALL THE BEST!!

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