Quick quick, help plzz

Mathematics

1 answer:

NemiM [27]2 years ago

5 0

Answer: The volume is 10 cm^3

Step-by-step explanation:

We know that each cube is a centimeter cube.

This means that if L is the length of each side of one of these cubes, then L = 1cm

And we know that the volume of a cube of side length L is:

V =L^3

Then the volume of each of the small cubes is:

V = (1 cm)^3 = 1 cm^3

Now we just need to count the number of cubes in the prism.

We can see that there is a "T" shape.

in the front part, we can see 5 cubes.

And there is another part in the back, where we also can assume that there are 5 cubes.

Then the total number of cubes in the prism is 10

And each of these cubes is 1 cm^3

Then the volume of the prism is 10 times 1 cm^3

The volume of the prism is:

V = 10*(1 cm^3) = 10 cm^3

You might be interested in

Help!! Algebra help will give stars

Inessa [10]

Answer:

I'll answer one question

Step-by-step explanation:

a. x >= -4. This reads, the domain is greater than and equal to negative four to infinity.

5 0

2 years ago

Let ρ = x3 + xe−x for x ∈ (0, 1), compute the center of mass.

hram777 [196]

The center of mass is mathematically given as

$\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}$

<h3>What is the center of mass.?</h3>

Determine the center of mass in one dimension:

Represent the masses at the respective distances.

$\begin{|c|c|} Masses \ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ Located at \\\rho=x^{3}+x \cdot e^{-x} & \ \ \ \ x \in(0,1)$ \\\end$

We calculate the total mass of the system.

$\begin{aligned}m &=\int_{0}^{1} \rho \cdot d x \\& m =\int_{0}^{1}\left(x^{3}+x \cdot e^{-x}\right) \cdot d x \\&m =\left|\frac{x^{4}}{4}-(x+1) e^{-x}\right|_{0}^{1} \\&m =\left(\frac{5}{4}-\frac{2}{e}\right)\end{aligned}$

Step 03: Calculate the moment of the system.

$\begin{aligned}M &=\int_{0}^{1}(\rho \cdot x) \cdot d x \\& M=\int_{0}^{1}\left(x^{4}+x^{2} \cdot e^{-x}\right) \cdot d x \\&M =\left|\frac{x^{5}}{5}-\left(x^{2}-2 x+2\right) \cdot e^{-x}\right|_{0}^{1} \\&M=\left(\frac{11}{5}-\frac{5}{e}\right)\end{aligned}$

we calculate the center of mass.

$\begin{aligned}\bar{x} &=\left(\frac{M}{m}\right) \\& \bar{x}=\left\{\left(\frac{\left.11-\frac{5}{5}\right)}{\left(\frac{5}{4}-\frac{2}{e}\right)}\right\}\right.\\& \bar{x}=\left(\frac{11 e-25}{5 e}\right) \cdot\left(\frac{4 e}{5 e-8}\right) \\&\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}$