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Gekata [30.6K]
3 years ago
11

The swimming pool at Spring Valley High School is a rectangle with a width of 75 meters and a length of 35 meters. Around the pe

rimeter of the pool is a tiled floor that extends w meters from the pool on all sides. Find an expression for the area of the tiled floor.
Mathematics
1 answer:
stealth61 [152]3 years ago
4 0
Same as before

total=(2w+75)(2w+35)=4w²+220w+2625

pool=75 times 35=2625

minus pool from total
4w²+220w+2625-2625=4w²+220w
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What is the circumference of a circle with a<br> diameter of 3 cm? (Use 3.14 as pi)
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circumference=9.42

Step-by-step explanation:

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Bill found a book he wanted on sale for $20.80. The original price of the book was $32. Find the discount rate.
Maurinko [17]

Answer:

35% discount

Step-by-step explanation:

We can first find what percent of 32 20.80 is.

This can be shown by using a percent proportion.

\frac{20.80}{32} = \frac{x}{100}

We can cross multiply to find the value of x.

20.80\cdot100=2080\\\\2080\div 32 = 65

So 20.80 is 65% of 32.

However, this is the percent of the original price. To find the discount, we have to subtract 65 from 100.

100 - 65 = 35

So the book was on a 35% discount.

Hope this helped!

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What is the simplest from of this expression? (-3x^2 + 4x) + (2x^2 - x -11)
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3 years ago
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
Galina-37 [17]

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

5 0
2 years ago
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