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3 years ago

The points (5, 3), (3, -4), (10, 3), and (8, -4) are the vertices of a polygon. What type of polygon is formed by these points?

Mathematics

2 answers:

Ulleksa [173]3 years ago

7 0

Answer:

The given coordinate is formed parallelogram.

B is correct.

Step-by-step explanation:

The points (5, 3), (3, -4), (10, 3), and (8, -4) are the vertices of a polygon.

Please see the attachment for point plot and name of coordinate

A(5,3) B(3,-4) C(8,-4) D(10,3)

First we find the slope of two points

$\text{Slope AB}=\dfrac{3+4}{5-3}=\dfrac{7}{2}$

$\text{Slope BC}=\dfrac{-4+4}{8-3}=0$

$\text{Slope CD}=\dfrac{4+3}{10-8}=\dfrac{7}{2}$

$\text{Slope AD}=\dfrac{3-3}{10-5}=0$

Slope of AB=Slope of CD=7/2

Slope of BC=Slope of AD=0

Opposite sides are parallel. Possible option is parallelogram.

Hence, The given coordinate is formed parallelogram.

zhuklara [117]3 years ago

3 0

Opposite sides are parallel. An appropriate choice is ...

B) Parallelogram

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3 years ago

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Solution :

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$r(t)=x(t)i+y(t)j$ is a parametrization of the most $\text{rapid increase curve}$ , then

$\frac{dx}{dt}=\frac{dx}{dt}i+\frac{dy}{dt}j$ is a tangent to the curve.

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$\frac{dy}{y}=2\frac{dx}{x}$

$\int\frac{dy}{y}=2\int \frac{dx}{x}$

$\ln y=2 \ln x$

⇒ y = kx^2 for some constant k

but we know that $y(x_0)=y_0$

$\Rightarrow kx_0^2=y_0$

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mariarad [96]

$\bf \sqrt{5x+19}-4=3x-15
\\\\
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\\\\
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\\\\
5x+19=(3x-11)^2\qquad 
\begin{cases}
(3x-11)^2\to (3x-11)(3x-11)\\
 --------------\\
\textit{or use FOIL to expand it}
\end{cases}$

then solve for "x"

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