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3 years ago

Is this graph a linear or no linear function?

Mathematics

2 answers:

drek231 [11]3 years ago

8 0

Answer:

Step-by-step explanation:

Linear equations are astaright line and have a constant rateof change

That line doesn't

olga2289 [7]3 years ago

5 0

it is nonlinear function

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When paid flat a round parachute has a diameter of 8 meters. What is the parachute’s circumference

mariarad [96]

Answer:

25.12 meters

Step-by-step explanation:

The circumference of a circle is given by

$c=2\pi r$ (1)

where:

$\pi\sim 3.14$

r is the radius of the circle

In this problem, the parachute has a diameter of 8 meters:

d = 8 m

The radius of a circle is equal to half of its diameter, so the radius of the parachute is:

$r=\frac{d}{2}=\frac{8}{2}=4 m$

And therefore, by applying eq.(1) we can find the circumference of the circle:

$c=2\pi r=2\cdot 3.14 \cdot 4 = 25.12 m$

7 0

3 years ago

Read 2 more answers

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.