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3 years ago

Can some one help me with this question its hard :( down below ill give brainliest

Mathematics

1 answer:

Alborosie3 years ago

6 0

Exponent law:
m^x * m^y = m^(x + y)
In other terms, when multiplying same bases, the exponents add
m^(2 + 2) = m^4

Solution: A. m^4

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Which statement about the following system of inequalities is true?

tatyana61 [14]

Answer: Option C) The solution contains points in three quadrants of the coordinate plane.

Please, see the attached files.

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8 0

3 years ago

Half the sum of a number and 1.75 is −5.1. Enter your answer as a simplified mixed number

musickatia [10]

The unknown number expressed as a mixed fraction is 11 19/20

Let the unknown number be "x"

The sum of the number and 1.75 is expressed as x + 1.75

Half the sum of a number and 1.75 is 1/2(x+1.75)

Equating the result to -5.1 wil give:

1/2(x+1.75) = -5.1

Solve the resulting equation

x + 1.75 = 2(-5.1)

x + 1.75 = -10.2

x = -10.2 - 1.75

x = 11.95

x = 1195/100

x = 11 95/100

Hence the unknown number expressed as a mixed fraction is 11 19/20

Learn more here: brainly.com/question/14583740

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.

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