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3 years ago

89.396668... rational or irrational number?

Mathematics

1 answer:

aksik [14]3 years ago

4 0

Answer:

89.396668... is a rational number

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tamaranim1 [39]

Answer:

m = v + 2

Step-by-step explanation:

m is Mags, and v is Vector

m (Mags' age) = v(Vector's age) + 2

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3 years ago

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Plz someone help on this question

sergey [27]

Answer:

Hi there!

Your answer is:

1050 liters per 6 days

Step-by-step explanation:

350liters per 2 days

*3 to determine 6 days

1050 liters per 6 days

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2 years ago

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vodomira [7]

Answer:

uumm

Step-by-step explanation:

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3 years ago

Find the slope of the line passing through the points (-4, -4) and (2, 5)

Sliva [168]

Answer: The slope is 3/2

Step-by-step explanation:

To find the slope, you need to find the difference between the y coordinates and divide it by the difference in the x coordinates.

The y coordinates are -4 and 5

The x coordinates are -4 and 2

-4 -5 = - 9

-4 - 2 = -6

-9/-6 = 3/2

4 0

3 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.

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