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miv72 [106K]
2 years ago
14

Lynne bought a bag of grapefruit, 1 5_ 8 pounds of apples, and 2 __3 16 pounds of bananas. The total weight of her purchases was

7 1_ 2 pounds. How much did the bag of grapefruit weigh?'
Mathematics
1 answer:
solmaris [256]2 years ago
3 0

Answer:

3 11/16 pounds

Step-by-step explanation:

Lynne bought a bag of grapefruit, 1 5/8 pounds of apples, and 2 3/16 pounds of bananas. The total weight of her purchases was 7 1/2 pounds. How much did the bag of grapefruit weigh?'

The Total weight = Weight of grape fruits + Weight of Apples + Weight of Bananas

Let the weight of the grape fruit = x

7 1/2 = x + 1 5/8 + 2 3/16

x = 7 1/2 - (1 5/8 + 2 3/16)

Lowest Common Denominator = 16

x= 7 1/2 - 3 + ( 10+ 3/16)

x = 7 1/2 - 3 + (13/16)

x = 7 1/2 - 3 13/16

Lowest Common Denominator = 16

x= 4 (8 - 13/16)

x = 4 (-5/16)

x = 3 11/16 pounds

Therefore, the bag of grapefruits weighed 3 11/16 pounds

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If y varies directly as x and y=32 when x=4. find the constant of variation​
Llana [10]

Answer:

k = 8

Step-by-step explanation:

Given that y varies directly as x then the equation relating them is

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To find k use the condition y = 32 when x = 4, then

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write the equation to solve for x 3x + 21 6x - 60 please help i was given 3x+21=6x-60 thats wrong can someone please help richar
Dmitry [639]

Answer:

<h2>3x + 21 = 6x - 60 </h2>

Step-by-step explanation:

parallel angles are equal

3x + 21 = 6x - 60

combine similar terms:

3x - 6x = -60 - 21

-3x = -81

x = 81/3

x = 27

now, plugin the value of x=27 back into the equation to get the angle:

= 3x + 21

= 3(27) + 21

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= 6x - 60

= 6(27) - 60

= 102

as you noticed, both values are equal 102 = 102 because its a parallel angles

 

so the equation is 3x + 21 = 6x - 60

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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

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