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Mathematics

1 answer:

storchak [24]2 years ago

7 0

Answer:

Step-by-step explanation:

The slope is: (y2 - y1) ÷ (x2-x1)

In this question y2 is 5, y1 is 8, x2 is 0, and x1 is -3.

5-8=-3

0-(-3)=3

-3 ÷ 3 = -1

the slope is negative 1

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Rachel purchased a prepaid phone card for $30. Long distance calls cost 6 cents a minute using this card. Rachel used her card o

Amiraneli [1.4K]

Answer: her call lasted for 26 minutes.

Step-by-step explanation:

Let x represent the number of minutes for which her call lasted.

Rachel purchased a prepaid phone card for $30. Long distance calls cost 6 cents a minute using this card. Converting 6 cents to dollars, it becomes 6/100 = $0.06

This means that the cost of x minutes of long distance call is

0.06 × x = $0.06x.

If the remaining credit on her card is $28.44, it means that

0.06x + 28.44 = 30

0.06x = 30 - 28.44

0.06x = 1.56

x = 1.56/0.06

x = 26

8 0

3 years ago

Any chance with 23-25?

iogann1982 [59]

This math answer is -2

4 0

4 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.

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