PLease help i will give brainliest

Mathematics

1 answer:

levacccp [35]2 years ago

4 0

Answer:

I don't have a good knowledge of means but I know it isn't B or D.

Step-by-step explanation:

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Evaluate:-19 - (-10) + (20) - (30) - (-15)<br> A) -4 <br> B) -3 <br> C) 3 <br> D) 4

Tresset [83]

So firstly, remember that <u>subtracting a negative number is the same as adding a positive number</u> so our expression will look like this:

$-19+10+20-30+15$

Now solve as such:

$\underbrace{-19+10}_{-9}+20-30+15\\\\\underbrace{-9+20}_{11}-30+15\\\\\underbrace{11-30}_{-19}+15\\\\\underbrace{-19+15}_{-4}\\\\-4$

<u>Your answer is -4, or A.</u>

4 0

3 years ago

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What is the name of a quadrilateral that has 4 equal sides but does not always have right angles

gayaneshka [121]

A quadrilateral with all sides equal is a rhombus.
If its angles are also all equal, then the rhombus
is a square.

5 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.

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