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Kruka [31]
3 years ago
9

Estimate The problem 2,731÷31

Mathematics
2 answers:
Vedmedyk [2.9K]3 years ago
4 0

Answer:

your answer should be 90

Step-by-step explanation:

Stolb23 [73]3 years ago
4 0

Answer:

90

Step-by-step explanation:

Start by rounding both numbers, 2700 divided by 30

giving you, 90

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Colin and Brian were playing darts. Colin scored 141. Brian scored 154 more than Colin. What was their combined score
creativ13 [48]

Answer:

436

Step-by-step explanation:

Colin’s score=141

Brian’s score= 141+154

=295

Total score = 295+141

=436

5 0
3 years ago
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7. Joey wants to find the area of the parallelogram below.
lesya692 [45]

Answer:

I'm not sure but I think it is b

3 0
3 years ago
Find the area of the following ellipse (round to nearest tenth).
goblinko [34]

Answer:

\huge\boxed{A=60\pi cm^2}

Step-by-step explanation:

The formula of an area of a elipse:

A=\pi ab

We have:

2a=10cm\to a=5cm\\\\2b=24cm\to b=12cm

Substitute:

A=\pi\dot5\cdot12=60\pi(cm^2)

8 0
2 years ago
Find the area of the figure.
sleet_krkn [62]

Answer:

17 units

Step-by-step explanation:

split the shape into several little ones that you have formulas for :)

6 0
3 years ago
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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
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
3 years ago
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