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3 years ago

Use the stem-and-leaf plot to answer the question.

Mathematics

2 answers:

Norma-Jean [14]3 years ago

7 0

Answer:

Step-by-step explanation:

mode is the most occurring number

dolphi86 [110]3 years ago

3 0

C 75 is the correct answer

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Choose the equation below that represents the line passing through the point (1, -- 4) with a slope of 1/2. (1 point)

Yakvenalex [24]

I think it’s the first one is not sure tho

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3 years ago

Draw a line representing the "run" and a line representing the "rise" of the line. State the slope of the line in simplest form.

Shtirlitz [24]

Answer:

$\displaystyle 1\frac{1}{3}$

Step-by-step explanation:

[3, 1] and [0, −3]

$\displaystyle \frac{-y_1 + y_2}{-x_1 + x_2} = m \\ \\ \frac{3 + 1}{0 + 3} = \frac{4}{3} = 1\frac{1}{3}$

I am joyous to assist you anytime.

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3 years ago

Read 2 more answers

I need help with two questions please, guathmath doesn't seem to help!

vivado [14]

The pink bar graph is 150

6 0

3 years ago

400 reduced by 3 times my age is 163

Blababa [14]

400 - 3x = 163

Add 3x to both sides.

400 = 163 + 3x

Subtract 163 to both sides.

237 = 3x

237/3 = 79

5 0

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