Between which pair of numbers is the exact product of 379 and 8 ?

A rectangular box has length 20cm, width 6cm and height 4cm. find how many cubes of size 2cm that will fit into the box.

Papessa [141]

Answer:

$\boxed{\pink{\sf The \ number \ of \ cubes \ that \ can \ be \ fitted \ is 60 .}}$

Step-by-step explanation:

Given dimensions of the box = 20cm × 6cm × 4cm .

Dimension of the cube = 2cm × 2cm × 2cm .

Therefore the number of cubes that can be fitted into the box will be equal to the Volume of box divided by the Volume of the cube. So ,

$\boxed{\red{\bf \implies No. \ of \ cubes \ = \dfrac{Volume \ of \ box}{Volume \ of \ cube }}}$

$\bf \implies n_{cubes} = \dfrac{20cm \times 6cm \times 4cm .}{2cm \times 2cm \times 2cm } \\\\\bf\implies n_{cubes} = 10 \ times 3cm \times 2cm \\\\\implies \boxed{\bf n_{cubes}= 60 }$

<h3>Hence the number of cubes that can be fitted in the box is 60 .</h3>

6 0

3 years ago

In which container is the probability of randomly selecting a green ball on one draw 4/7?

Snowcat [4.5K]

Answer:

B

Step-by-step explanation:

B

3 0

3 years ago

Read 2 more answers

GIVING BRAINILEST-

Nutka1998 [239]

Answer:

To construct a dilation, we need the center of dilation, which is the point from which we make the image smaller or larger. We also need a scale factor, which is the number that determines how large or small we are making the new image. If the scale factor is greater than 1, we have an enlargement, because the new image will be larger than the original image. If the scale factor is less than 1, we have a reduction because the new image will be smaller than the original image.

Enlargement

Step-1

Let's dilate triangle PQR by a scale factor of 2 with a center of dilation at point C.

Step-2

First, we will use a straightedge to connect the center of dilation C to each vertex. We are going to extend those lines across the whole page.

Step-3

Next, we'll place the point of the compass on the center of dilation C and the pencil on vertex Q and measure that distance.

Step-4

Now, without changing the size of the compass, we will move the point of the compass to vertex Q, and make a mark on the line that is extended through Q.

Step-5

Notice that since we measured the distance from C to Q and then used that same distance to mark the line, we have now created a new point that is twice the distance from C that Q was.

We now repeat the process for each of the other vertices of the triangle.

Step-6

These marks represent the vertices of our dilated image. We often use the same letters with an apostrophe to show that it is the corresponding vertex.

Step-7

Finally, we use the straightedge to connect all of the new vertices.

Now,See how triangle P

′

Q

′

R

′

is twice the size of the original triangle.

Reduction

Step-1

Given triangle PQR, now let's dilate the image with a center of dilation at C and with a scale factor of

2

1

Step-2

Using the straightedge, we will connect all of the vertices to the center of dilation C. In this case, since our scale factor is less than 1, we are reducing the size of the triangle.

Step-3

The green lines are the new lines that we have drawn.

Since we have to reduce the size of the triangle by half, we want to cut the distance in half from each vertex to the center of dilation. In order to do this, we will set our compass to a distance that is more than halfway between vertex P and point C but less than the total distance.

Step-4

Keeping the compass at that same distance, start with the point of the compass on vertex P, then draw a curve that intersects the line between P and C. Next, keeping the compass at the same distance, place the point of the compass on C and draw a curve that intersects the line between P and C and intersects the previously drawn curve.

Step-5

Now, we will take the straightedge and connect the points where the curves overlap.