What is the area of an equalateral triangle of side length x?

Mathematics

1 answer:

OLEGan [10]2 years ago

5 0

First you need to make a line that is perpendicular to one side. The diagram is shown below. then the length of the line that is perpendicular to the triangle is
$\frac{ \sqrt{3}x }{2}$
then the area of the equilateral triangle equals to
$\frac{1}{2} height \times base$
so the area would be
$\frac{ \sqrt{3} {x}^{2} }{4}$

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wolverine [178]

If it has no stretch factor, then there will be no coefficient. Meaning this is a lot easier. So, we equation is

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20 points and brainliest <br> I’m in quiz in need it asap <br> Number 4

iren [92.7K]

Answer and step-by-step explanation:

The polar form of a complex number $a+ib$ is the number $re^{i\theta}$ where $r = \sqrt{a^2+b^2}$ is called the modulus and $\theta = tan^-^1 (\frac ba)$ is called the argument. You can switch back and forth between the two forms by either remembering the definitions or by graphing the number on Gauss plane. The advantage of using polar form is that when you multiply, divide or raise complex numbers in polar form you just multiply modules and add arguments.

(a) let's first calculate moduli and arguments

$r_1 = \sqrt{(-2\sqrt3)^2+2^2}=\sqrt{12+4} = 4\\ \theta_1 = tan^-^1(\frac{2}{-2\sqrt3}) =-\pi/6\\r_2=\sqrt{1^2+1^2}=\sqrt2\\ \theta_2 = tan^-^1(\frac 11)= \pi/4$

now we can write the two numbers as

$z_1=4e^{-i\frac \pi6}; z_2=e^{i\frac\pi4}$

(b) As noted above, the argument of the product is the sum of the arguments of the two numbers:

$Arg(z_1\cdot z_2) = Arg(z_1)+Arg(z_2) = -\frac \pi6 + \frac \pi4 = \frac\pi{12}$

(c) Similarly, when raising a complex number to any power, you raise the modulus to that power, and then multiply the argument for that value.

$(z_1)^1^2=[4e^{-i\frac \pi6}]^1^2=4^1^2\cdot (e^{-i\frac \pi6})^1^2=2^2^4\cdot e^{-i(12)\frac\pi6}\\=2^2^4 e^{-i\cdot2\pi}=2^2^4$

Now, in the last step I've used the fact that $e^{i(2k\pi+x)} = e^i^x ; k\in \mathbb Z$ , or in other words, the complex exponential is periodic with $2\pi$ as a period, same as sine and cosine. You can further compute that power of two with the help of a calculator, it is around 16 million, or leave it as is.