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Liono4ka [1.6K]
2 years ago
15

I am super bad at graphing. The Statement says "Graph each system and determine the number of solutions that it is. Tell if the

system has One Solution, Infinitely Many Solutions, or No Solutions. If it has one solution, name it using an ordered pair. Also tell if the system is consistent or inconsistent, and tell if the system is independent or dependent, or neither"
The Systems are:
x + 2y = 6
3x + 6y = 8​
Mathematics
2 answers:
erma4kov [3.2K]2 years ago
5 0

Answer:

get that degree girl

Step-by-step explanation:

focus on me girl

Aleks [24]2 years ago
3 0

Answer: to be honest i don’t know  hope this helps

Step-by-step explanation:good luck

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Hope I helped!
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Answer: y=2x+3

Step-by-step explanation:

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2 years ago
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.

7 0
1 year ago
What is the measure of XYZ 36 Degrees <br> Please help
exis [7]
B is the correct answer




good luck
8 0
2 years ago
What is the solution to the inequality a-3&gt;5
Lady bird [3.3K]

Answer:

a > 8

Step-by-step explanation:

a-3>5

Add 3 to each side

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a > 8

6 0
3 years ago
Read 2 more answers
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