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3 years ago

Hmmmmmmmmmmmmmmmmmmmmmmmmmmm

Mathematics

1 answer:

belka [17]3 years ago

4 0

$\dfrac{6}{7}cd=\dfrac{6}{7}\cdot c\cdot d\\\\Factors:\ \dfrac{6}{7},\ c,\ d\\\\6x^2+7xy+3+9\\\\Constans\ (without\ x\ and\ y):\ 3,\ 9\\\\2x-4y+8=(2x)+(-4y)+(8)\\\\Terms:\ 2x,\ -4y,\ 8$

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HELP ASAP!!!! EXPLAIN

Amanda [17]

Answer:

i think the answer is 50

Step-by-step explanation:

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3 years ago

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Find the median of the set of data this box-and-whisker plot represents.

egoroff_w [7]

Answer:

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Step-by-step explanation:

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Graph △XYZ with vertices X(2, 3), Y(−3, 2), and Z(−4,−3) and its image after the translation (x, y)→(x+3, y−1)

viktelen [127]

Answer:

X'(5, 2)
Y'(0, 1)
Z'(-1, -4)

Step-by-step explanation:

The translation increases each x-coordinate by 3, moving the point 3 units to the right. It decreases each y-coordinate by 1, moving the point 1 unit down.

(x, y) ⇒ (x+3, y-1)

X(2, 3) ⇒ X'(5, 2)

Y(-3, 2) ⇒ Y'(0, 1)

Z(-4, -3) ⇒ Z'(-1, -4)

The red arrows show the translation of each point in the graph.

5 0

3 years ago

Select from the following choices those graphs that represent only a dilation applied to this polygon with a scale factor great

romanna [79]

Answer:

B and C

Step-by-step explanation:

Required

Select graphs that are dilated by a scale factor greater than 1

For graph A:

Graph A is smaller than the original graph. This indicates dilation with a scale factor less than 1

For graph B:

Graph B is bigger than the original graph and is dilated over (0,0). This indicates dilation with a scale factor greater than 1

For graph C:

Graph C is bigger than the original graph; however, it is not dilated over (0,0). This indicates dilation with a scale factor greater than 1

For graph D:

Graph D is bigger than the original graph; however, it is not only dilated but also flipped over (i.e. rotated).

<em>Hence, b and c is true</em>

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3 years ago

If we inscribe a circle such that it is touching all six corners of a regular hexagon of side 10 inches, what is the area of the

Brrunno [24]

Answer:

$\left(100\pi - 150\sqrt{3}\right)$ square inches.

Step-by-step explanation:

<h3>Area of the Inscribed Hexagon</h3>

Refer to the first diagram attached. This inscribed regular hexagon can be split into six equilateral triangles. The length of each side of these triangle will be $10$ inches (same as the length of each side of the regular hexagon.)

Refer to the second attachment for one of these equilateral triangles.

Let segment $\sf CH$ be a height on side $\sf AB$ . Since this triangle is equilateral, the size of each internal angle will be $\sf 60^\circ$ . The length of segment

$\displaystyle 10\, \sin\left(60^\circ\right) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$ .

The area (in square inches) of this equilateral triangle will be:

$\begin{aligned}&\frac{1}{2} \times \text{Base} \times\text{Height} \\ &= \frac{1}{2} \times 10 \times 5\sqrt{3}= 25\sqrt{3} \end{aligned}$ .

Note that the inscribed hexagon in this question is made up of six equilateral triangles like this one. Therefore, the area (in square inches) of this hexagon will be:

$\displaystyle 6 \times 25\sqrt{3} = 150\sqrt{3}$ .

<h3>Area of of the circle that is not covered</h3>

Refer to the first diagram. The length of each side of these equilateral triangles is the same as the radius of the circle. Since the length of one such side is $10$ inches, the radius of this circle will also be $10$ inches.

The area (in square inches) of a circle of radius $10$ inches is:

$\pi \times (\text{radius})^2 = \pi \times 10^2 = 100\pi$ .

The area (in square inches) of the circle that the hexagon did not cover would be:

$\begin{aligned}&\text{Area of circle} - \text{Area of hexagon} \\ &= 100\pi - 150\sqrt{3}\end{aligned}$ .

3 0

3 years ago