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snow_lady [41]
3 years ago
9

I WILL GIVE BRAINLIEST, 5 STARS, AND THANKS!

Mathematics
1 answer:
elixir [45]3 years ago
6 0

Answer:

D. 27

Step-by-step explanation:

1.5 cm / 0.5 cm = 3

So now think of the cube as 3 blocks x 3 blocks x 3 blocks, instead of 1.5cm x 1.5cm x 1.5cm. 3 x 3 x 3 = 27

I hope this helped!

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Nathan and Steven stacked boxes on a shelf . Nathan lifted 13 boxes and Steven lifted 15 boxes . The boxes that Nathan lifted ea
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The answer is N=13(x+12), where x represents the amount of pounds Steven's boxes weighed and N represents the total number of pounds Nathan lifted. 
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Answer for Solve similars triangles advanced solve for x
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Step-by-step explanation:

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Find the area of the figure
Eduardwww [97]

Answer:

132cm²

Step-by-step explanation:

You can find the area of the entire shape by calculating the area of seperate smaller shapes.

You can split them up into 2 rectangles, one with a base of 7cm and height of 4cm, and another one with a base of 7cm and height of 8cm. You also have 2 triangles, with a base of 19cm - 7cm / 2 = 6cm since it is the entire bottom line subtracted by the top; they will have a height of 8cm each.

To calculate the rectangles areas, simply do length x width, which are:

4cm x 7cm = 28cm²

8cm x 7cm = 56cm²

Next, since the triangles are similar in measurements, instead of doing the normal calculation to find the area of a triangle, you can do base times height again, which is:

6cm x 8cm = 48cm²

Add all the answers together to get the final area, which is 132cm².

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Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 21 in. by 12 in. by
a_sh-v [17]

Answer:

Dimension of the box is 16.1\times 7.1\times 2.45

The volume of the box is 280.05 in³.

Step-by-step explanation:          

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 21 in. by 12 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the box?

Solution :

Let h be the height of the box which is the side length of a corner square.

According to question,

A sheet of cardboard 21 in. by 12 in. by cutting congruent squares from the corners and folding up the sides.

The length of the box is L=21-2h

The width of the box is W=12-2h

The volume of the box is V=L\times W\times H

V=(21-2h)\times (12-2h)\times h

V=(21-2h)\times (12h-2h^2)

V=252h-42h^2-24h^2+4h^3

V=4h^3-66h^2+252h

To maximize the volume we find derivative of volume and put it to zero.

V'=12h^2-132h+252

0=12h^2-132h+252

Solving by quadratic formula,

h=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

h=\frac{-(-132)\pm\sqrt{132^2-4(12)(252)}}{2(12)}

h=\frac{132\pm72.99}{24}

h=2.45,8.54

Now, substitute the value of h in the volume,

V=4h^3-66h^2+252h

When, h=2.45

V=4(2.45)^3-66(2.45)^2+252(2.45)

V\approx 280.05

When, h=8.54

V=4(8.54)^3-66(8.54)^2+252(8.54)

V\approx -170.06

Rejecting the negative volume as it is not possible.

Therefore, The volume of the box is 280.05 in³.

The dimension of the box is

The height of the box is h=2.45

The length of the box is L=21-2(2.45)=16.1

The width of the box is W=12-2(2.45)=7.1

So, Dimension of the box is 16.1\times 7.1\times 2.45

6 0
4 years ago
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