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Minchanka [31]
3 years ago
9

Find the area ratio of a cube with volume 125 m to a cube with volume 64 m.

Mathematics
2 answers:
Viefleur [7K]3 years ago
5 0
V=s^3
SA=6s^2
so

125=v
125=s^3
cube root both sides
5=s
A=6(5)^2=6(25)=150 m^2


64=v
64=s^3
cube root both sides
4=s

A=6(4)^2=6(16)=96 m^2

ratio is

150:96=25:16
Over [174]3 years ago
4 0
So to find the ratio of the *Surface areas you need to find the Area of the sides/faces of the cubes.

*Keep in mind that the faces of a cube are squares.

Surface Area of a cube = Area of a side * 6
Area = length * width
Area of a square = length * length 

We don't know the length/width of the cube, so we need to solve for it using the values for the volumes of the two cubes.

Volume = length * width* height\\Volume of a cube = length * length * length\\\\125\ m^{3}=length^{3}\\\\\sqrt[3]{125\ m^{3}}=\sqrt[3]{length^{3}}\\\\5\ m=length\\\\\\64\ m^{3}=length^{3}\\\\\sqrt[3]{64\ m^{3}}=\sqrt[3]{length^{3}}\\\\4\ m=length


Now find the area of the sides:

Area = length *length\\\\Area\ of\ cube\ 1=5\ m*5\ m=25\ m^{2}\\\\Surface\ Area\ of\ cube\ 1=side\ area*6\\\\SA_{cube\ 1}=25\ m^{2}*6=150\ m^{2}\\\\\\Area\ of\ cube\ 2=4\ m*4\ m=16\ m^{2}\\\\Surface\ Area\ of\ cube\ 2=side\ area*6\\\\SA_{cube\ 2}=16\ m^{2}*6=96\ m^{2}


Now find the ratio between the Surface areas:
150\ m^{2}:96\ m^{2}\\\\150\6:96\6\\\\25:16=\frac{25}{16}=1\ \frac{9}{16}=1.5625


Thus the answer is 25/16 = 1.5625
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Answer:

A. The football does not reach a height of 15ft

Step-by-step explanation:

Given

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Required

Determine which of the options is true

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To solve this, we make use of maximum of a function

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i.e we first solve for \frac{-b}{2a}

Then substitute \frac{-b}{2a} for x in f(x) = ax^2 + bx + c

In our case:

First we need to solve \frac{-b}{2a}

Then substitute \frac{-b}{2a} for t in h(t) = -16t^2 + 30t

By comparison:

b = 30

c = -16

\frac{-b}{2a} = \frac{-30}{2 * -16}

\frac{-b}{2a} = \frac{-30}{-32}

\frac{-b}{2a} = \frac{30}{32}

\frac{-b}{2a} = \frac{15}{16}

Substitute \frac{15}{16} for t in h(t) = -16t^2 + 30t

h(\frac{15}{16}) = -16(\frac{15}{16})^2 + 30(\frac{15}{16})

h(\frac{15}{16}) = -16(\frac{225}{256}) + \frac{450}{16}

h(\frac{15}{16}) = -(\frac{225}{16}) + \frac{450}{16}

h(\frac{15}{16}) = \frac{-225 + 450}{16}

h(\frac{15}{16}) = \frac{225}{16}

h(\frac{15}{16}) = 14.0625

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Answer:

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Step-by-step explanation:

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Length of AD = 3 units

By applying Pythagoras theorem in ΔDBE,

DE² = DB² + BE²

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DE = √18

DE = 4.24 units

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By applying Pythagoras theorem in ΔABC,

AC² = AB² + BC²

AC² = 6² + 6²

AC = √72

AC = 8.49 units

Perimeter of ADEC = 3 + 4.24 + 3 + 8.49

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Area of ADEC = Area of ΔABC - Area of ΔBDE

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Area of ADEC = 18 - 4.5

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Step-by-step explanation:

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Answer:

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