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1 year ago

Given angle ABC shown on the coordinate plane below.

Mathematics

1 answer:

Crank1 year ago

6 0

The attached image shows the image of A"B"C" after the transformation

<h3>What is the transformation of the triangle about?</h3>

The transformation rule states:

A"B"C" = Ro90° (T(-4,3)(ABC))

This implies that one need to rotate the triangle in a 90⁰ clockwise direction, and then one need to translate the triangle.

Using the image shown, the coordinates of ABC are;

A = (-1, 2)

B = (1, 4)

C = (3, -1)

The 90⁰ rule clockwise rotation will be:

(x,y) -- (y,-x)

So, when translated, it will be:

A' = (2, 1)

B' = (4, -1)

C' = (-1, -3)

Then the translation of the triangle using T(-4,3):

(x, y) - (x - 4, y + 3)

So, there is:

A'' = (-2, 4)

B'' = (0, 2)

C'' = (-5, 0)

Learn more about transformation from:

brainly.com/question/28108536

#SPJ1

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3 years ago

An 8-sided die is formed when two square pyramids are attached by their bases. The perimeter of the base shape is 6 centimeters.

lawyer [7]

Answer:

7.5cm²

Complete question:What is the surface area of the 8-sided figure?

Step-by-step explanation:

Consider the half of the 8-sided die i.e the pyramid ABCDE:

As the base is 6 of the perimeter, so each side of the base is 6/4=3/2

Now, the distance between the tallest points is 2 cm, so the height of Pyramid ABCDE is 1 cm.

Considering the right triangle EOM:

EO=1, OM= 3/4 (it is half of AB=3/2) and EM is the hypothenuse, so by the pythagorean theorem length of EM is

$\sqrt{1+ ( \frac{3}{4} )^{2} }= \sqrt{1+ \frac{9}{16} }= \sqrt{ \frac{25}{16} }= \frac{5}{4} (cm)$

5. So side EBC has area

$\frac{1}{2}BC*EM= \frac{1}{2}* \frac{3}{2}* \frac{5}{4}= \frac{15}{16}$

6. The total area is 8*Area(EBC)= $8* \frac{15}{16}= \frac{15}{2}=7.5 ( cm^{2} )$

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3 years ago

Find in degrees the numeric value of the acute angle of rotation that eliminates the product term from the equation 7x2+24xy−34x

lutik1710 [3]

Rotating the graph of $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ with $B\ne0$ counterclockwise by $\theta$ eliminates the $xy$ term, where $\cot2\theta=\frac{A-C}{B}.$ Plugging in, we have $\cot2\theta=\frac{7}{24},$ since $C=0.$ Solving, we have $\theta=\frac{1}{2}\cot^{-1}(\frac{7}{24})\approx\boxed{36.9^\circ}.$