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Lana71 [14]
3 years ago
11

Ameeta finished 12 math problems. David finished 3 times as many. How many more math problems did David finish than Ameeta?

Mathematics
1 answer:
Volgvan3 years ago
8 0
  • Answer:
  • david finished 24 problems more than ameetha
  • Step-by-step explanation:
  • <em>problems finished by ameetha = 12 problems</em>
  • <em>problems finished by ameetha = 12 problemsproblems finished by David = 3 times more than ameethas problem = 3*12 = 36</em>
  • <em>problems finished by ameetha = 12 problemsproblems finished by David = 3 times more than ameethas problem = 3*12 = 36the problems finished by David more than ameetha =</em>
  • <em>problems finished by ameetha = 12 problemsproblems finished by David = 3 times more than ameethas problem = 3*12 = 36the problems finished by David more than ameetha =36-12= 24 </em>
  • <em>problems finished by ameetha = 12 problemsproblems finished by David = 3 times more than ameethas problem = 3*12 = 36the problems finished by David more than ameetha =36-12= 24 so David finished 24 problems more than ameetha</em>
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x=-7

Step-by-step explanation:

subtract 3 from both sides

-6x=45-3

-6x=42

divide each term by 6

\frac{-6x}{-6} =\frac{42}{-6}

x=\frac{42}{-6}

simplify

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What is the smallest rotation clockwise about the center of the figure that will carry the entire figure onto itself.
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Step-by-step explanation:

A regular n-sided polygon has n axes of symmetry.  The angle between two consecutive axes is 360° / n.

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rhombus and a square have one and the same side of 6 cm. The area of the rhombus is 4/5 of the area of the square. Find the heig
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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

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