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Sliva [168]
3 years ago
13

Consider the following literal equation.

Mathematics
1 answer:
galina1969 [7]3 years ago
4 0

We can solve this in many ways, let's try this one.

T∧2 = 4 pi∧2 * a∧3 / GM    First we will multiply whole equation with variable( or parameter) M and get

M * T∧2 = 4 pi∧2 * a∧3 / G  After that we will divide whole equation with variable ( or parameter )  T∧2 and get

M = 4 pi∧2 * a∧3 / G * T∧2

This is correct answer

Good luck!!!

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Step-by-step explanation:

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3 years ago
Tower one, a cube with a hexagonal prism measures 15.6cm. Tower two, a cube with a cylinder measures 18.3cm. Tower 3, a hexagona
lina2011 [118]

Answer:

Tower 4 = 23.9cm

Step-by-step explanation:

Given

Tower one = 15.6 cm

Tower two = 18.3 cm

Tower 3 = 13.9 cm.

Required:

Height of the 4th tower

Represent a cube by X; a cylinder by Y and a hexagonal prism by Z

Tower one, a cube with a hexagonal prism = X + Z = 15.6

Tower two, a cube with a cylinder = X + Y = 18.3

Tower 3, a hexagonal prism with a cylinder = Z + Y = 13.9

X + Z = 15.6 ----- Equation 1

X + Y = 18.3 ----- Equation 2

Z + Y = 13.9 ----- Equation 3

Subtract equation 1 from 2

(X + Y = 18.3) - (X + Z = 15.6)

X - X + Y - Z = 18.3 -15.6

Y - Z = 18.3 -15.6

Y - Z = 2.7 ---- Equation 4

Add Equation 4 to Equation 3

(Y - Z = 2.7) + (Z + Y = 13.9)

Y + Y - Z + Z = 2.7+ 13.9

2Y  = 2.7+ 13.9

2Y  = 16.6

Divide both sides by 2

\frac{2Y}{2}  = \frac{16.6}{2}

Y  = \frac{16.6}{2}

Y  = 8.3cm

Substitute Y  = 8.3cm in Equation 2 and 3

X + Y = 18.3 ----- Equation 2

X + 8.3 = 18.3

Subtract 8.3 from both sides

X + 8.3 - 8.3 = 18.3 - 8.3

X = 18.3 - 8.3

X = 10cm

Z + Y = 13.9 ----- Equation 3

Z + 8.3 = 13.9

Subtract 8.3 from both sides

Z + 8.3 - 8.3 = 13.9 - 8.3

Z = 13.9 - 8.3

Z = 5.6cm

So, we have that

X = 10cm

Y  = 8.3cm

Z = 5.6cm

The question states that the 4th tower is made up of the three shapes;

This implies that;

Tower 4 = X + Y + Z

Tower 4 = 10cm + 8.3cm + 5.6cm

Tower 4 = 23.9cm

The height of the 4th tower is 23.9cm

8 0
3 years ago
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