All categories

4 years ago

The completed construction of a regular hexagon is shown below. Explain why ACF is a 30º-60º-90º triangle.

Mathematics

1 answer:

LuckyWell [14K]4 years ago

5 0

We know that one regular hexagon is consist of the six equilateral triangles.

One of this is ΔAFM every angle in this triangle worth 60°, according to this

∡AFM=60°.

ΔABC is isosceles triangle with angle ∡ABC=120°, because every interior angle in regular hexagon worth 120°.

Angles ∡BAC=∡BCA=30°. Total sum of angles in one triangle is 180°.

∡ACF≅∡BCA=30° at the end we can conclud that angle ∡FAC=90° WHY?

Good luck!!!

You might be interested in

What is the value of x?<br> Enter your answer in the box.

Lynna [10]

That my friend is number 6

3 0

3 years ago

Please answer this correctly

Leona [35]

Answer:

The first answer.

Step-by-step explanation:

The first answer.

What was the temperature of the air?

'was 2C warmer than the surface of the ice'

warmer than= +2c

Temp. of ice=-2c

-2c+2c=0

7 0

3 years ago

Read 2 more answers

Simplify the following leaving your answer in positive index<br><br> 1) (x/y)^-4 x (y/x)^8

Basile [38]

Answer:

$(\frac{x}{y})^{-4} * (\frac{y}{x})^8 = (\frac{y}{x})^{12$

Step-by-step explanation:

Given

$(\frac{x}{y})^{-4} * (\frac{y}{x})^8$

Required

Determine the solution to this

$(\frac{x}{y})^{-4} * (\frac{y}{x})^8$

In law of incides:

$\frac{a}{b} = \frac{b}{a}^{-1}$

This implies that:

$(\frac{x}{y})^{-4} * (\frac{y}{x})^8$ is equivalent to:

$(\frac{y}{x})^{-(-4)} * (\frac{y}{x})^8$

$(\frac{y}{x})^{4} * (\frac{y}{x})^8$

Apply first law of indices:

$(\frac{y}{x})^{4+8$

$(\frac{y}{x})^{12$

Hence:

$(\frac{x}{y})^{-4} * (\frac{y}{x})^8 = (\frac{y}{x})^{12$

5 0

3 years ago

7p + 8p -12 = 59 cans someone help me on this please

notka56 [123]

Combine like terms
7p+8p-12=59
15p-12=59
add 12 to both sides
15p+12-12=59+12
15p+0=71
15p=71
divide both sides by 15
p=4 and 11/15

7 0

4 years ago

Finding Derivatives Implicity In Exercise, find dy/dx implicity.<br> x2 - 3 ln y + y2 = 10

Veseljchak [2.6K]

Answer:

$\frac{dy}{dx}=-\frac{2xy}{2y^2-3}$

Step-by-step explanation:

We are given that

$x^2-3lny+y^2=0$

Differentiate w.r.t x

$2x-\frac{3}{y}\frac{dy}{dx}+2y\frac{dy}{dx}=0$

By using formula

$\frac{dx^n}{dx}=nx^{n-1}$

$\frac{d(lnx)}{dx}=\frac{1}{x}$

$\frac{dy^n}{dx}=ny^{n-1}\frac{dy}{dx}$

$\frac{dy}{dx}(-\frac{3}{y}+2y)+2x=0$

$\frac{dy}{dx}(-\frac{3}{y}+2y)=-2x$

$\frac{dy}{dx}=-\frac{2x}{-\frac{3}{y}+2y}$

$\frac{dy}{dx}=-\frac{2xy}{2y^2-3}$

Hence, the derivative of function

$\frac{dy}{dx}=-\frac{2xy}{2y^2-3}$

8 0

4 years ago