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GarryVolchara [31]
3 years ago
10

16 Which coordinate plane represents the linear relationship y = 2x + 4

Mathematics
1 answer:
borishaifa [10]3 years ago
8 0

Answer:

the answer is D

Step-by-step explanation:

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Find the product of z1 and z2, where z1 = 7(cos 40° + i sin 40°) and z2 = 6(cos 145° + i sin 145°).
Oduvanchick [21]
For two complex numbers z_1=re^{i\theta}=r(\cos\theta+i\sin\theta) and z_2=se^{i\varphi}=s(\cos\varphi+i\sin\varphi), the product is

z_1z_2=rse^{i(\theta+\varphi)}=rs(\cos(\theta+\varphi)+i\sin(\theta+\varphi))

That is, you multiply the moduli and add the arguments. You have z_1=7e^{i40^\circ} and z_2=6e^{i145^\circ}, so the product is

z_1z_2=7\times6(\cos(40^\circ+145^\circ)+i\sin(40^\circ+145^\circ)=42(\cos185^\circ+i\sin185^\circ)=42e^{i185^\circ}
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3 years ago
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What is the greatest perfect square of 1290
Alex73 [517]
<span>It has no perfect square factors, unless you count 1 = 1^2. That's sort of a degenerate case we don't usually count, since every integer has that factor. We would usually say that 1290 is a square-free integer. Hope this helps!!</span>
6 0
2 years ago
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PLEASE HELP ASAP THANK YOU
masha68 [24]
1-85.71% of the team attendance
6 0
3 years ago
Point P'P
Snowcat [4.5K]

<u>Given</u>:

The point P' is the image of the point P under the translation (x,y) \implies (x-6, y-1)

The coordinates of the point P are (6,0)

We need to determine the coordinates of the point P'

<u>Coordinates of the point P':</u>

The coordinates of the point P' can be determined by substituting the coordinates of the point P(6,0) in the translation.

Thus, substituting the coordinates, we have;

(6,0)\implies (6-6,0-1)

Simplifying the coordinates, we get;

(6,0)\implies (0,-1)

Thus, the coordinates of the point P' is (0,-1)

5 0
3 years ago
Nine more than half of a number is 21
pogonyaev
I assume you're looking for the number:
0.5n + 9 = 21
- 9
0.5n = 12
÷ 0.5
n = 24
So the beginning number is 24, hope this helps!
3 0
3 years ago
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