X + x + 2 = 14
2x + 2 = 14
2x = 12, x = 6
Solution: the smallest of the two integers is 6
In the given you wrote -2, whereas it should be -2i
z = a + bi and in trigo form | z |.(cos Ф + i.sin Ф).
z = 0 -2i → | z |.(cos Ф + i.sin Ф)
Now let's calculate z:
z² = a²+b² → z² = 0² (-2.i)² → z = -2(i)² →z= -2(-1) → z = |2|
tan Ф = b/a = -2/0 → tan Ф → - ∞ ↔ Ф = -90° or Ф = 270°
then the - 2 I ↔ |2|(cos 270° + i.sin 270°)
First we need to determine what the 6 angles must add to. Turns out we use this formula
S = 180(n-2)
where S is the sum of the angles (result of adding them all up) and n is the number of sides. In this case, n = 6. So let's plug that in to get
S = 180(n-2)
S = 180(6-2)
S = 180(4)
S = 720
The six angles, whatever they are individually, add to 720 degrees. The six angles are y, y, 2y-20, 2y-20, 2y-20, 2y-20, <span>
They add up and must be equal to 720, so let's set up the equation to get...
(y)+(y)+(</span>2y-20)+(2y-20)+(2y-20)+(<span>2y-20) = 720
Let's solve for y
</span>y+y+2y-20+2y-20+2y-20+2y-20 = 720
10y-80 = 720
10y-80+80 = 720+80
<span>10y = 800
</span>
10y/10 = 800/10
y = 80
Now that we know the value of y, we can figure out the six angles
angle1 = y = 80 degrees
<span>angle2 = y = 80 degrees
</span><span>angle3 = 2y-20 = 2*80-20 = 140 degrees
</span>angle4 = 2y-20 = 2*80-20 =<span> 140 degrees
</span><span>angle5 = 2y-20 = 2*80-20 = 140 degrees
</span>angle6 = 2y-20 = 2*80-20 =<span> 140 degrees
</span>
and that's all there is to it
I suppose I can do that and end up spending 6.50$?
1.25 x 5.20 = 6.50 final amount due.
Answer:
The best population for Maya's school would be all the seventh grade students at Maya's school, because that is who she wants to know how many games they play.
D. All the seventh grade students at Maya's school.
Hope this helps ;)
