All categories

3 years ago

Dimension of a cuboid is 4 m 3 m ad 5 m how many such cuboid can form a cube....... please and urgent..experts please help

Mathematics

1 answer:

larisa [96]3 years ago

4 0

Step-by-step explanation:

We dimension of the smallest cube to be made from cuboids of sides 3 m , 4 m and 5 m will be the least common multiple of 3 m, 4 m and 5 m i.e. 60 m

12 cuboidal should be stacked along 5 m edge to 60 m, 15 cuboids should be stacked along 4 m edge and 20. cuboids should be stacked along edge to make a cube of 60 m edge, Hence number of cuboids are 12× 15 ×20=3600

hope it helps

You might be interested in

I need help please this is for math

maxonik [38]

5x + 13 Hope this helps you

8 0

3 years ago

Graph the equation y=-X – 2 using the intercepts.

ahrayia [7]

Answer:

y intercept = (0,-2)

x intercept = (-2,0)

Step-by-step explanation:

For y intercept, plug 0 for x

y = (0) - 2

y = -2

y intercept = -2

For x intercept, plug 0 for y

(0) = -x - 2

Move -2 to other side

-x = 2

Since x is negitave, make 2 negitave

x = -2

x intercept = -2

4 0

2 years ago

The bucket is 30% full. when half way empty it is 15% full. how full is the bucket when 250% full?

Sunny_sXe [5.5K]

Answer:

125%

Step-by-step explanation:

divide 250% by 2

7 0

2 years ago

Read 2 more answers

Find x so that the points (x,x+1), (x+2,x+3) and (x+3,2x+4) form a right-angled triangle.

azamat

Let a, b, and c be vectors each starting at the origin and terminating at the points (x, x + 1), (x + 2, x + 3), and (x + 3, 2x + 4), respectively.

Then the vectors a - b, a - c, and b - c are vectors that point in directions parallel to each of the legs formed by the triangle with these points as its vertices.

If this triangle is to contain a right angle, then exactly one of these pairs of vectors must be orthogonal. In other words, one of the following must be true:

(a - b) • (a - c) = 0

or

(a - b) • (b - c) = 0

or

(a - c) • (b - c) = 0

We have

a - b = (x, x + 1) - (x + 2, x + 3) = (-2, -2)

a - c = (x, x + 1) - (x + 3, 2x + 4) = (-3, -x - 3)

b - c = (x + 2, x + 3) - (x + 3, 2x + 4) = (-1, -x - 1)

Case 1: If (a - b) • (a - c) = 0, then

(-2, -2) • (-3, -x - 3) = (-2)×(-3) + (-2)×(-x - 3) = 2x + 12 = 0 ==> x = -6

which would make a - c = (-3, 3) and b - c = (-1, 5), and their dot product is not zero. Then the triangles vertices are at the points (-6, -5), (-4, -3), and (-3, -8).

Case 2: If (a - b) • (b - c) = 0, then

(-2, -2) • (-1, -x - 1) = (-2)×(-1) + (-2)×(-x - 1) = 2x + 4 = 0 ==> x = -2

which would make a - c = (-3, -1) and b = (-1, 1), and their dot product is also not zero. The vertices are the points (-2, -1), (0, 1), and (1, 0).

Case 3: If (a - c) • (b - c) = 0, then

(-3, -x - 3) • (-1, -x - 1) = (-3)×(-1) + (-x - 3)×(-x - 1) = x ² + 4x + 6 = 0

but the solutions to x here are non-real, so we throw out this case.

So there are two possible values of x that make a right triangle, x = -6 and x = -2.

3 0

3 years ago

Which is the value of this expression when a=5 and k=-2

Zinaida [17]

Answer:

Option C is correct.

Step-by-step explanation:

We are given the expression:

$(\frac{3^2a^{-2}}{3a^{-1}})^k$

The value of a =5 and k = -2

Putting the values and solving

$=(\frac{3^2*5^{-2}}{3*5^{-1}})^-2\\=(\frac{3^{2-1}}{5^{-1+2}})^-2\\=(\frac{3^{1}}{5^{1}})^-2\\\\=(\frac{3}{5})^-2\\if \,\,a^{-1} \,\,then\,\, 1/a\\=\frac{(3)^{-2}}{(5)^{-2}}\\ Can\,\,be\,\,written\,\,as\\\\=\frac{(5)^{2}}{(3)^{2}} \\=\frac{25}{9}$

Option C is correct.

3 0

3 years ago