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3 years ago

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What is the value of -4+15(18÷0.5)

1 answer:

AveGali [126]3 years ago

6 0

<span><span>−4</span>+<span>15<span>(<span>18/0.5</span>)</span></span></span><span>=<span><span>−4</span>+<span><span>(15)</span><span>(36)</span></span></span></span><span>=<span><span>−4</span>+540</span></span><span>=536</span> <span> </span>

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Step-by-step explanation:

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A punch glass is in the shape of a hemisphere with a radius of 5 cm. If the punch is being poured into the glass so that the cha

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Answer:

28.27 cm/s

Step-by-step explanation:

Though Process:

The punch glass (call it bowl to have a shape in mind) is in the shape of a hemisphere
the radius $r=5cm$
Punch is being poured into the bowl
The height at which the punch is increasing in the bowl is $\frac{dh}{dt} = 1.5$
the exposed area is a circle, (since the bowl is a hemisphere)
the radius of this circle can be written as $'a'$

what is being asked is the rate of change of the exposed area when the height $h = 2 cm$
the rate of change of exposed area can be written as $\frac{dA}{dt}$ .
since the exposed area is changing with respect to the height of punch. We can use the chain rule: $\frac{dA}{dt} = \frac{dA}{dh} . \frac{dh}{dt}$

and since $A = \pi a^2$ the chain rule above can simplified to $\frac{da}{dt} = \frac{da}{dh} . \frac{dh}{dt}$ -- we can call this Eq(1)

Solution:

the area of the exposed circle is

$A =\pi a^2$

the rate of change of this area can be, (using chain rule)

$\frac{dA}{dt} = 2 \pi a \frac{da}{dt}$ we can call this Eq(2)

what we are really concerned about is how $a$ changes as the punch is being poured into the bowl i.e $\frac{da}{dh}$

So we need another formula: Using the property of hemispheres and pythagoras theorem, we can use:

$r = \frac{a^2 + h^2}{2h}$

and rearrage the formula so that a is the subject:

$a^2 = 2rh - h^2$

now we can derivate a with respect to h to get $\frac{da}{dh}$

$2a \frac{da}{dh} = 2r - 2h$

simplify

$\frac{da}{dh} = \frac{r-h}{a}$

we can put this in Eq(1) in place of $\frac{da}{dh}$

$\frac{da}{dt} = \frac{r-h}{a} . \frac{dh}{dt}$

and since we know $\frac{dh}{dt} = 1.5$

$\frac{da}{dt} = \frac{(r-h)(1.5)}{a}$

and now we use substitute this $\frac{da}{dt}$ . in Eq(2)

$\frac{dA}{dt} = 2 \pi a \frac{(r-h)(1.5)}{a}$

simplify,

$\frac{dA}{dt} = 3 \pi (r-h)$

This is the rate of change of area, this is being asked in the quesiton!

Finally, we can put our known values:

$r = 5cm$

$h = 2cm$ from the question

$\frac{dA}{dt} = 3 \pi (5-2)$

$\frac{dA}{dt} = 9 \pi cm/s// or//\frac{dA}{dt} = 28.27 cm/s$

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3 years ago