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maksim [4K]
3 years ago
14

Find the coordinates of the midpoint of HK if H(-1,2) and K(-7,-4)

Mathematics
2 answers:
devlian [24]3 years ago
5 0
The coordinates would be at (-4.5,-1)
tankabanditka [31]3 years ago
5 0

Answer:

The midpoint of HK is (-4, 1)

Step-by-step explanation:

We are given two coordinates H(-1, 2) and K(-7, -4)

We need to find the midpoint of HK.

We have formula to find the midpoint.

(\frac{x_1 + x_2}{2} , \frac{y_1 + y_2}{2} }

Here x_1 = -1, y_1 = 2, x_2 = -7 and y_2 = -4

Now plug in these values in the above formula, we get

(\frac{-1 + (-7)}{2} , \frac{2 +(-4)}{2} )

(\frac{-8}{2}, \frac{-2}{2}  ) =  (-4, -1)

So, the midpoint of HK is (-4, 1)

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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}

Read more about the center of mass.

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