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3 years ago

A cube has a side length of 6 centimeters what is the volume

Mathematics

2 answers:

IgorC [24]3 years ago

5 0

Answer:

216cm³

Step-by-step explanation:

Semenov [28]3 years ago

3 0

Answer:

the answer is 24

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Please answer thanks a lot!

Evgen [1.6K]

Vol (pyr) = 1/3 b × h, where h = 15 and b = base = area of triangular base = 1/2 b×h, where h = 12 and b = 13
V = 1/3 (1/2×12×13)×15
V = (1/3×1/2×12)×13×15
V = 2×13×15 = 30×13 = 390 in^3

6 0

4 years ago

Y=-x-4 <br>y=x<br><br>solve system of equations algebraically

Serhud [2]

Answer:

y=-2 x=-2

Step-by-step explanation:

y=-x-4

y=x

you can plug y in for x in the first equation by using substitution

y=-y-4

and then you solve for y

2y=-4

y=-2

so y equals -2. Since y=x, x also equals -2. Therefore

y=-2 x=-2

4 0

3 years ago

The second side of a triangular deck is 4 feet longer than the shortest side. The third side that is 4 feet shorter than twice t

Svetlanka [38]

The second side of a triangular deck is 4 feet longer than the shortest side

(s+4) = the 2nd side

and a third side that is 4 feet shorter than twice the length of the shortest side.

(2s-4) = the 3rd side

If the perimeter of the deck is 48 feet, what are the lengths of the three sides?

s + (s+4) + (2s-4) = 48

Combine like terms

s + s + 2s + 4 - 4 = 48

4s = 48

s = 48/4

s = 12 ft is the shortest side

I'll let you find the 2nd and 3rd sides, ensure they add up to 48

Hope this helps!

5 0

3 years ago

A right circular cylinder is inscribed in a sphere with diameter 4cm as shown. If the cylinder is open at both ends, find the la

SOVA2 [1]

Answer:

$8\pi\text{ square cm}$

Step-by-step explanation:

Since, we know that,

The surface area of a cylinder having both ends in both sides,

$S=2\pi rh$

Where,

r = radius,

h = height,

Given,

Diameter of the sphere = 4 cm,

So, by using Pythagoras theorem,

$4^2 = (2r)^2 + h^2$ ( see in the below diagram ),

$16 = 4r^2 + h^2$

$16 - 4r^2 = h^2$

$\implies h=\sqrt{16-4r^2}$

Thus, the surface area of the cylinder,

$S=2\pi r(\sqrt{16-4r^2})$

Differentiating with respect to r,

$\frac{dS}{dr}=2\pi(r\times \frac{1}{2\sqrt{16-4r^2}}\times -8r + \sqrt{16-4r^2})$

$=2\pi(\frac{-4r^2+16-4r^2}{\sqrt{16-4r^2}})$

$=2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})$

Again differentiating with respect to r,

$\frac{d^2S}{dt^2}=2\pi(\frac{\sqrt{16-4r^2}\times -16r + (-8r^2+16)\times \frac{1}{2\sqrt{16-4r^2}}\times -8r}{16-4r^2})$

For maximum or minimum,

$\frac{dS}{dt}=0$

$2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})=0$

$-8r^2 + 16 = 0$

$8r^2 = 16$

$r^2 = 2$

$\implies r = \sqrt{2}$

Since, for r = √2,

$\frac{d^2S}{dt^2}=negative$

Hence, the surface area is maximum if r = √2,

And, maximum surface area,

$S = 2\pi (\sqrt{2})(\sqrt{16-8})$

$=2\pi (\sqrt{2})(\sqrt{8})$

$=2\pi \sqrt{16}$

$=8\pi\text{ square cm}$

4 0

3 years ago

Solve for t. P = irt

Hoochie [10]

A. and ir ≠ 0 .............

6 0

3 years ago

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