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o-na [289]
2 years ago
10

Using the image above, if the density of the LEGO piece is 0.85 g/cm^3, what COULD be a possible measurement for the density of

oil
0.79 g/cm^3

0.8599 g/cm^3

0.95 g/cm^3

1.75 g/cm^3

Mathematics
1 answer:
grin007 [14]2 years ago
6 0

The Lego is submerged under the oil and therefore, is more dense than the oil

The possible density of the oil is the option;

<u>0.79 g/cm³</u>

Reason:

Given:

The density of Lego = 0.85 g/cm³

Density of oil = Required

Solution:

The density of an object is the mass per unit volume it occupies

  • Density = \dfrac{Mass}{Volume}

Increase in density increases the mass a unit volume occupies, therefore, a

more dense object is heavier than a less dense object such that the less

dense object will float on a dense object

The oil is floating on the Lego, therefore, the Lego is more dense than the

oil, and we have;

  • Density of Lego > Density of oil

The density of Lego = 0.85 g/cm³

Therefore;

0.85 g/cm³ > Density of oil

From the given options, the option that has a density less than 0.85 g/cm³ is the the option 0.79 g/cm³

Therefore, the possible density of the oil is <u>0.79 g/cm³</u>

<u />

Learn more here:

brainly.com/question/12864420

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Answer:

179.159

Step-by-step explanation:

Given the sequence :

50, 60, 72,..

The sequence given is a geometric sequence :

For a geometric sequence ;

ar^(n - 1)

a = first term

r = a2 / a1 = 60 / 50 = 1.2

The 8th term ; n =8

a8 = ar^(n - 1)

a8 = 50*1.2^(8 - 1)

a8 = 50*1.2^7

a8 = 50*3.5831808

a8 = 179.15904

a8 = 179.159 (nearest thousandth )

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3 years ago
jenna's rectangular garden borders a wall. she buys 80 ft of fencing. what are the dimensions of the garden that will maximize i
FrozenT [24]

Answer:

The  dimensions are x =20 and y=20 of the garden that will maximize its area is 400

Step-by-step explanation:

Step 1:-

let 'x' be the length  and the 'y' be the width of the rectangle

given Jenna's buys 80ft of fencing of rectangle so the perimeter of the rectangle is    2(x +y) = 80

                         x + y =40

                               y = 40 -x

now the area of the rectangle A = length X width

                                                  A = x y

substitute 'y' value in above A = x (40 - x)

                                              A = 40 x - x^2 .....(1)

<u>Step :2</u>

now differentiating equation (1) with respective to 'x'

                                      \frac{dA}{dx} = 40 -2x     ........(2)

<u>Find the dimensions</u>

<u></u>\frac{dA}{dx} = 0<u></u>

40 - 2x =0

40 = 2x

x = 20

and y = 40 - x = 40 -20 =20

The dimensions are x =20 and y=20

length = 20 and breadth = 20

<u>Step 3</u>:-

we have to find maximum area

Again differentiating equation (2) with respective to 'x' we get

\frac{d^2A}{dx^2} = -2

Now the maximum area A =  x y at x =20 and y=20

                                        A = 20 X 20 = 400

                                         

<u>Conclusion</u>:-

The  dimensions are x =20 and y=20 of the garden that will maximize its area is 400

<u>verification</u>:-

The perimeter = 2(x +y) =80

                           2(20 +20) =80

                              2(40) =80

                              80 =80

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P = Present value

R = interest rate

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