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3 years ago

Can yall help me????

Mathematics

1 answer:

Kryger [21]3 years ago

3 0

Hi There!

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Remember to follow PEMDAS.

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Problem #1:

= (25 - 6.25) + (5 · 3)²

= (18.75) + (15)²

= 18.75 + 225

= 243.75

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Problem #2:

= (8 + 3)² + 10 · (-4)

= (11²) + 10 · (-4)

= 121 + 10 · (-4)

= 121 + (-40)

= 81

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Hope This Helps :)

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4 years ago

The radius of a cone is increasing at a constant rate of 7 meters per minute, and the volume is decreasing at a rate of 236 cubi

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$\frac{dV}{dt} = \frac{1}{3}\pi \frac{d}{dt} (uh)$

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$\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h\frac{du}{dt}]$

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Using the Chain's rule

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Then,

$\frac{dV}{dt} = \frac{1}{3}\pi [r^{2} \frac{dh}{dt} + h(2r) \frac{dr}{dt}]$

Now,

From the question

$\frac{dr}{dt} = 7 m/min$

$\frac{dV}{dt} = 236 m^{3}/min$

At the instant when $r = 99 m$

and $V = 180 m^{3}$

We will determine the value of h, using

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

$180 = \frac{1}{3}\pi (99)^{2}h$

$180 \times 3 = 9801\pi h$

$h =\frac{540}{9801\pi }$

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$236 = \frac{1}{3}\pi [(99)^{2} \frac{dh}{dt} + (\frac{20}{363\pi }) (2(99)) (7)]$

$236 \times 3 = \pi [9801 \frac{dh}{dt} + (\frac{20}{363\pi }) 1386]$

$708 = 9801\pi \frac{dh}{dt} + \frac{27720}{363}$

$708 = 30790.75 \frac{dh}{dt} + 76.36$

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$\frac{dh}{dt}= \frac{631.64}{30790.75}$

$\frac{dh}{dt} = 0.021 m/min$

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