What is the slope of the line below?

Mathematics

2 answers:

Vlad1618 [11]2 years ago

5 0

Answer:

3/2

Step-by-step explanation:

do rise over run.

pogonyaev2 years ago

3 0

3/2 rise over run....

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A sequence of two transformations that can be used to show that polygon ABCDE is congruent to polygon GFJIH

Burka [1]

Two transformations that can be used to show that polygon ABCDE is congruent to polygon GFJIH : <u>rotation -90° and translation (0,2)</u>

<h3>Further explanation</h3>

There are several forms of transformation, including translation, dilation, rotation or reflection

There are 2 shapes of geometry:

Similar figures:

the same shape but different in size,

Congruent figures:

the same size and the same shape

Both polygons will appear congruent if, after being transformed, the two polygons will map exactly to each other

A sequence of two transformations e can do is :

1. rotation -90°

Let point H(4,1) from polygon GFJIH , we rotate -90°

$\left[\begin{array}{ccc}x'\\y'\end{array}\right]=\left[\begin{array}{ccc}0&1\\-1&0\end{array}\right]\left[\begin{array}{ccc}4-6\\1+1\end{array}\right]+\left[\begin{array}{ccc}6\\-1\end{array}\right]\\\\=\left[\begin{array}{ccc}0&1\\-1&0\end{array}\right]\left[\begin{array}{ccc}-2\\2\end{array}\right]+\left[\begin{array}{ccc}6\\-1\end{array}\right]\\\\=\left[\begin{array}{ccc}8\\1\end{array}\right]$

2. translation (0,2)

$\left[\begin{array}{ccc}8\\1\end{array}\right]+\left[\begin{array}{ccc}0\\2\end{array}\right]=\left[\begin{array}{ccc}8\\3\end{array}\right]$

<h3 /><h3>Learn more</h3>

brainly.com/question/13715226

Keywords: transformation, translation, rotation

#LearnWithBrainly

7 0

2 years ago

Read 2 more answers

Solve the inequality 2x>30+5/4x

insens350 [35]

Answer:

Step-by-step explanation:

$2x > 30+\frac{5}{4x} \\2x-\frac{5}{4x} > 30\\\frac{8x^2-5}{4x} > 30\\case~1\\if~x > 0\\8x^2-5 > 120x\\8x^2-120x > 5\\x^2-15x > \frac{5}{8} \\adding~(-\frac{15}{2} )^2~to~both~sides\\(x-\frac{15}{2} )^2 > \frac{5}{8}+\frac{225}{4} \\(x-\frac{15}{2} )^2 > \frac{455}{8} \\x-\frac{15}{2} < -\sqrt{\frac{455}{8} } \\x < \frac{15}{2}-\sqrt{\frac{455}{8} } \\or~x < 0\\rejected~as~x > 0$

$x-\frac{15}{2} > \sqrt{\frac{455}{8} } \\x > \frac{15}{2} +\sqrt{\frac{455}{8} }$

case~2

$if~x < 0\\8x^2-5 < 120x\\8x^2-120x < 5\\x^2-15x < \frac{5}{8} \\adding~(-\frac{15}{2} )^2\\(x-\frac{15}{2} )^2 < \frac{5}{8} +(-\frac{15}{2} )^2\\|x-\frac{15}{2} | < \frac{5+450}{8} \\-\sqrt{\frac{455}{8} } < x-\frac{15}{2} < \sqrt{\frac{455}{8} } \\\frac{15}{2} -\sqrt{\frac{455}{8} } < x < \frac{15}{2} +\sqrt{\frac{455}{8} } \\but~x < 0\\7.5-\sqrt{\frac{455}{8} } < x < 0$