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2 years ago

How to find out the slope

1 answer:

6 0

To find the slope, you divide the difference of the y-coordinates of 2 points on a line by the difference of the x-coordinates of those same 2 points .

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Through any 2 points there is exaclty_____line.

mr Goodwill [35]

Answer:

One

Step-by-step explanation:

Through any 2 points, there is exactly one line.

6 0

2 years ago

Read 2 more answers

Given that APOR is similar to APTS, what is the value of x?

marusya05 [52]

Answer:

Step-by-step explanation:

8 0

2 years ago

The new club sticker is in the shape of a triangle. It has a height of 8 inches and a base of 10 inches. The area of the club st

rusak2 [61]

Answer:

40 square inches.

Step-by-step explanation:

It is given that the new club sticker is in the shape of a triangle.

Height of triangle = 8 inches

Base of triangle = 10 inches

We need to find the area of sticker.

Area of triangle $=\dfrac{1}{2}\times base \times height$

Substitute base = 10 and height = 8 in the above formula.

Area of triangle $=\dfrac{1}{2}\times 10\times 8$

Area of triangle $=\dfrac{80}{2}$

Area of triangle $=40$

Therefore, the area of the club sticker is 40 square inches.

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.

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