All categories

4 years ago

Line GH contains points (-2,6) and H (5,-3). What is the slope of GH

Mathematics

2 answers:

svetlana [45]4 years ago

8 0

Answer:

-9/7

Step-by-step explanation:

To find the slope given 2 points, we use the formula

m = (y2-y1)/(x2-x1)

where (x1,y2) and (x2,y2) are the two points

m = (-3-6)/(5--2)

m = (-3-6)/(5+2)

= -9/7

SOVA2 [1]4 years ago

8 0

Answer:

y=-1.3x+3.4

Step-by-step explanation:

You might be interested in

What is the length of a side of the cube shown with given volume?

tekilochka [14]

｡☆✼★ ━━━━━━━━━━━━━━━ ☾

Cube root the value:

Answer = 9 ft

If you work forward from this answer, you'll understand the process

volume = l x w x h

volume = 9 x 9 x 9

volume = 729

Have A Nice Day ❤

Stay Brainly! ヅ

- Ally ✧

｡☆✼★ ━━━━━━━━━━━━━━━━ ☾

7 0

3 years ago

Write an algebraic expression for the phrase the product of 4 and s.

notka56 [123]

Answer: 4s

Step-by-step explanation:

Product means multiply

4 times s = 4s

8 0

3 years ago

Read 2 more answers

(6+3)+21=6+(3+21) name each property of addition or multiplication

Anna007 [38]

(6 + 3) + 21 = 6 + (3 + 21)

It's an ASSOCIATIVE PROPERTY
(a + b) + c = a + (b + c)

7 0

4 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

Solve for f. Show your work! 8=f-(13-2)

Nana76 [90]

Answer:

So F=19

Step-by-step explanation:

8=f-(13-2)

8=f-11

+11 +11 You plus 11 by both sides

19=f

You can check you answer by plugging it in

8=19-(13-2)

your answer would still be 19

8 0

3 years ago