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Veronika [31]
3 years ago
8

PLZ HELP DUE IN 2 MINS 

Mathematics
1 answer:
Zepler [3.9K]3 years ago
8 0
Sanakwekeiiehrdnmdmdldodd
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What is 645 round to the nearest ten
madreJ [45]
The answer is 650 have a nice day
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Help pls, Which graph correctly shows point A representing z1 − z2 on a complex plane where z1 and z2 are complex numbers?
Marianna [84]

Answer:

The first diagram is the correct one

Step-by-step explanation:

Notice that the subtraction of two complex numbers (z1- z2) implies the use of the opposite for the real and imaginary part of the complex number that is subtracted (in our case of z2). When we do such, the complex number z2 gets reflected about the origin (0,0), and then the real components of the two numbers get added among themselves and the imaginary components get added among themselves.

The diagram that shows such reflection about the origin [ z2 = 3 + 5 i being converted into -3 - 5 i] and then the combination of real parts [-3 + 5 = 2] and imaginary parts [-5 i - 3 i = - 8 i], is the very first diagram shown.

4 0
3 years ago
Devin's Crackers will make 7,536 ounces of cheese crackers next year. The company plans to put the crackers into 48-ounce boxes.
Arturiano [62]

Answer:

157 boxes

Step-by-step explanation:

Take the total amount of ounces to be made, and divide by the 'size' of the box. with 7,536 ounces to be made and the boxes being 48 ounces: 7536/48 will equal 157 boxes filled.

5 0
3 years ago
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How do I do this explain
xeze [42]
So you would take 3x1+2 that gives you 5 and you would keep the same denominator and so that would give you 5/3 then do the same to the other side which would give you 5/4 then you would times it and that would give you 2 1/12
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Integrate [0, 1/2] xcos(pi*x).
elena-14-01-66 [18.8K]
To answer your question, this could be the possible answer and i hope you understand and interpret it correctly:
<span>[Integrate [0, 1/2] xcos(pi*x
let u=x so that du=dx
 and v=intgral cos (xpi)dx
        v=(1/pi)sin(pi*x)
 integration by parts
uv-itgral[0,1/2]vdu just plug ins
 (1/pi)sinpi*x]-(1/pi)itgrlsin(pi*x)dx from 0 to 1/2
(1/pi)x sinpi*x - (1/pi)[-(1/pi) cos pi*x] from 0 to 1/2
 =(1/2pi)+(1/pi^2)[cos pi*x/2-cos 0]
=1/2pi - 1/2pi^2
=(pi-2)/2pi^2 ans</span>
3 0
3 years ago
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