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Ivanshal [37]
2 years ago
5

Find the geometric mean of 6 and 48

Mathematics
1 answer:
r-ruslan [8.4K]2 years ago
5 0

Answer:

48+6= 54

54/2= 27

Answer is 27

Step-by-step explanation:

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10 pounds of sugar for 5 litres of cream.
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Regular triangular pyramid has 6 cm long base edge and slant height k=9 cm. Find the lateral area of the pyramid.
gregori [183]

Answer:

81 cm²

Step-by-step explanation:

Since, the lateral face of a triangular pyramid is a triangle,

Given,

The base edge or the base of one lateral face of pyramid, a = 6 cm,

And, the slant height or the height of the face, k = 9 cm,

Thus, the area of one lateral face of the pyramid,

A=\frac{1}{2}\times a\times k

=\frac{1}{2}\times 6\times 9

=\frac{54}{2}

=27\text{ square cm}

We know that, a Regular triangular pyramid has 3 lateral faces,

Hence, the total lateral area of the pyramid,

L.A.=3\times\text{ The area of one lateral face}

=3\times 27

=81\text{ square cm}

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2 years ago
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3 years ago
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Solve the equations. Make conclusion. 3|x+12|=6<br><br> *Please put down the steps also.*
Blababa [14]
3 | x+12 |=6
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3 0
3 years ago
Determine the point of intersection of right bisectors in a triangle ∆ with vertices A (-3, 5), B (1, 1) and (−7, −3). Find the
iris [78.8K]

Answer:

Point of intersection (-11/3 , 1/3)

All distances from the vertices are \frac{10\sqrt{2} }{3}

Step-by-step explanation:

The vertices of the triangle is A(-3,5) , B (1,1) and C (-7,-3)

We need to find the perpendicular bisector of the triangle first

let's take one side connecting A and B

mid point of A and B = (-1,3)

Slope of the line joining A and B = -1

slope of perpendicular to line joining A and B = 1

equation of line passing through (-1,3) with slope 1

y - 3 = 1(x-(-1))

y -3 = x+1

x-y = -4 ............(1)

similarly

mid point joining B and C = (-3,-1)

slope perpendicular to line joining B and C  =  -2

Equation of perpendicular bisector of line joining B and C =

y +1 = -2(x +3 )

y+1 = -2x -6

2x+y = -7 ..........(2)

On solving 1 and 2

x= -11/3 , y= 1/3

Distances

From A = \frac{10\sqrt{2} }{3}

From B = \frac{10\sqrt{2} }{3}

From C = \frac{10\sqrt{2} }{3}

7 0
2 years ago
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