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laiz [17]
3 years ago
11

If z = 3cis30°, z3 in rectangular form is _____ +_____ i.

Mathematics
1 answer:
oee [108]3 years ago
6 0
The rectangular form is 3c/30=z
Z=10 3c//30
10z+70x + 17x-10z
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AB is parallel to CD. EG = FG ˆ AEG = 110° ˆ Calculate the size of DGH.
Sever21 [200]

Opposite exterior angles are congruent.

DGH = AEG

DGH = 110

3 0
4 years ago
(2a^x +1/2 b^x)^2 thank you!!!!!!!!!!!!!!!!!!!1
kogti [31]
Note that (a+b)^2 = a^2 + 2ab + b^2.

Therefore, 
                                                                              b^x         (b^x )^2
<span>(2a^x +1/2 b^x)^2 becomes (2a^x)^2 + 2(2a^x)(-------- ) + ------------
                                                                                 2               4

LCD is 4.  Thus, we have
                                             
            4 [ (2a^(2x) ] + 8</span>a^x*b^x + b^(2x)
<span>                                  </span>
3 0
4 years ago
Find the volume of the pyramid.
lana66690 [7]

Answer:

140

Step-by-step explanation:

B=35

1/3 (35)(12)

= 140

6 0
3 years ago
Irections: Simplify each expressio<br> 25-(√16-1)-(3-9)²
enyata [817]

Answer:

-14

Step-by-step explanation:

\sqrt{16} = 4\\\\(\sqrt{16} - 1) = (4 -1) =3\\\\(3-9) = (-6)\\\\(3-9) ^2 = (-6)^2 = 36\\\\25 -  3 - 36 = 25 - 39 =  -14\\

3 0
2 years ago
One option in a roulette game is to bet $12 on red.​ (There are 18 red​ compartments, 18 black​ compartments, and two compartmen
Alona [7]

Answer:

The expected value is -$0,63

Step-by-step explanation:

The expected value of a discrete variable x is given by:

E(x)=x1*p(x1)+x2*p(x2)+...+xn*p(xn)

Where x1, x2, .. xn are the posibles results that the variable x can take and p(x1), p(x2),...,P(xn) are their respective probabilities.

In this case, we have two possible results if we bet $12 on red. This results are:

1. The balls lands on red: in this case we keep the $12 we paid and we awarded 12. That means, we gain $12 and the probability of this case is the division of the number of red compartments by total number of compartments. So:

x1=$12

p(x1)=18/38

2. The balls doesn't land on red: in this case we lose the $12 that we bet and the probability is the division of number of compartments that aren't red by the total number of compartments. So:

x2=-$12

p(x2)=20/38

Then, replacing x1, x2, p(x1) and p(x2) on the equation of the expected value, we get:

E(x)=($12)*(18/38)  +  (-$12)*(20/38)

E(x)=$-0.63

So, the expected value if we bet $12 on red is to lose $0.63

5 0
4 years ago
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