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Nookie1986 [14]
3 years ago
9

Find the volume of a cube with side lengths of fifteen feet

Mathematics
1 answer:
lidiya [134]3 years ago
8 0

Answer:

3375 feet cube

Step-by-step explanation:

Given :

side of cube = 15 feet

Formula used :

Volume of cube , v = a^3

Solution :

Volume of cube

= a^3

= 15^3

= 15*15*15

= 3375 feet cube

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If x is 32 mm what’s the area of the triangle
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Answer:

4608*sqrt(3)

Step-by-step explanation:

x + 2x = 3x, so 3x is the height of the triangle.  This is a 90-60-30 triangle so the base would be 3x*sqrt(3).  To find the area, we can multiply the base with the height and divide by two, so it would be (3x)(3x*sqrt(3))/2 or (9/2)x^2 * sqrt(3).  x = 32, and plugging this into the equation gets us to (9/2)32^2 * sqrt(3), which is 4608*sqrt(3).  

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. A triangle has vertices on a coordinate grid at points J(-1, 5), K(4, 5), and L(4, -2). What is the length, in units, of ?
Neko [114]

Answer:

20.6

Step-by-step explanation:

Given data

J(-1, 5)

K(4, 5), and

L(4, -2)

Required

The perimeter of the traingle

Let us find the distance between the vertices

J(-1, 5) amd

K(4, 5)

The expression for the distance between two coordinates is given as

d=√((x_2-x_1)²+(y_2-y_1)²)

substitute

d=√((4+1)²+(5-5)²)

d=√5²

d= √25

d= 5

Let us find the distance between the vertices

K(4, 5), and

L(4, -2)

The expression for the distance between two coordinates is given as

d=√((x_2-x_1)²+(y_2-y_1)²)

substitute

d=√((4-4)²+(-2-5)²)

d=√-7²

d= √49

d= 7

Let us find the distance between the vertices

L(4, -2) and

J(-1, 5)

The expression for the distance between two coordinates is given as

d=√((x_2-x_1)²+(y_2-y_1)²)

substitute

d=√((-1-4)²+(5+2)²)

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d=  √74

d=8.6

Hence the total length of the triangle is

=5+7+8.6

=20.6

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2 years ago
Angle MNO is formed by segments MN and NO on the following coordinate grid:
tekilochka [14]

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