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pav-90 [236]
3 years ago
6

HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
2 answers:
LekaFEV [45]3 years ago
8 0
It’s A!!! Lol dneneenne
bezimeni [28]3 years ago
7 0
Pick a pick a dont actually
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What is the relationship between polar coordinates, vectors, complex numbers
Leno4ka [110]

Answer:

Answer: There is also a connection between multiplication of complex numbers and polar coordinates. As usual, polar coordinates of a point ( x , y ) in the plane are defined as ( r , θ ) where r is the length of the vector ( x , y ) and θ is the angle this vector makes with the positive x axis.

hope this helps

Step-by-step explanation:

4 0
3 years ago
Minnie has 4 different stamps. She needs 3 to mail a large letter. How many ways can she pick the 3 that will be used?
timurjin [86]
Number of ways of picking a smaller number out of a bigger number is called combination.
Number of way of picking 3 stamps out of 4 is 4C3 (4 combination 3) = 4! / (3! x (4 - 3)!) = 4! / (3! x 1!) = (4 x 3 x 2 x 1) / (3 x 2 x 1 x 1) = 24 / 6 = 4 ways.
7 0
3 years ago
Integrate the following by parts |x^2 e^2x dx​
Travka [436]

Answer:

The answer is shown in the picture. you basically have to do integration by parts 2 times to get final answer.

7 0
3 years ago
-4.3&lt;-2.91&lt;5.7 <br>is that that true?
IRISSAK [1]
Yes that is correct
7 0
3 years ago
Can someone PLEASE explain how to simplify square roots with variables and eexponents in them?? I'd also be thankful if you expl
Verizon [17]
X^a/b is  \sqrt[b]{x^a} . The way I memorise that is x^1/3 is the cubic root of x. Do you get it? In that case, x is raised to a power of 1 and the cubic root is practically has a power of 3.
In your example, 

\sqrt[ \frac{3}{2} ]{16 x^4} is practically square rooting each term then cubing them individually. Remember when square-rooting any index you halve it. I'll elaborate:

\sqrt{x^4} = x^{2}
\sqrt{16} = 4
Then cube each,
4^3 = 64 
and ( x^{2} )^3 = x^{6}

As for the 2nd part: you must use the rules of indices.
x^{a}  *  x^{b} =  x^{a+b}
So breaking the question up:

3 * 3 = 9
x^{ \frac{1}{2} } stays as is since the 2nd term does not contain x
now: 
y^{ \frac{4}{3} }  * y^{1} =  y^{ \frac{4}{3} + 1 }  =  y^{ \frac{4}{3} +  \frac{3}{3} }  =   y^{ \frac{7}{3} }
This makes your final answer look like this:
9 x^{ \frac{1}{2} }  y^{ \frac{7}{3} }

I hope that helped and good luck in your test!
4 0
4 years ago
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