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2 years ago

1: 18 is __ % of 30

Mathematics

1 answer:

Zigmanuir [339]2 years ago

4 0

Answer:

1. 60%

2. 70

3. 8%

Step-by-step explanation:

I attached pictures to show my work (sorry if they’re a bit messy)!

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Plzz hurry<br> 12/19(x-22)=38-6x-3

nadya68 [22]

Answer:

x = 929/126

Step-by-step explanation:

12/19(x-22)=38-6x-3

Combine like terms

12/19(x-22)=35-6x

Multiply each side by 19

19* 12/19(x-22)=19*(35-6x)

12( x-22) = 19*(35-6x)

Distribute

12x - 264 = 665 - 114x

Add 114x to each side

12x +114x - 264 = 665 - 114x+114x

126x -264 = 665

Add 264 to each side

126x-264+264 = 665+264

126x = 929

Divide each side by 126

126x/126 = 929/126

x = 929/126

4 0

3 years ago

Read 2 more answers

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.

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