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kotykmax [81]
4 years ago
12

Hey! please help me posted picture of question:)

Mathematics
1 answer:
Ilia_Sergeevich [38]4 years ago
6 0
The answer is a and c

Hope this helps.
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∣1−9∣ ÷∣−8−8∣=<br> Answer as a fraction.
Goshia [24]

Answer:

The answer is \frac{1}{2}

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Factor the trinomial: 3x2 + 11x + 6
OleMash [197]
3x^2+11x+6

(3x^2+2x) + (9x+6)

Factor out x  from 

3x^2+2x  =  x(3x+2)

Factor out 3  from

9x+6 = 3(3x+2)

= x(3x+2)+3(3x+2)

Factor out common  term (3x+2) :

=(3x+2)(x+3)

hope this helps!

5 0
3 years ago
Find the distance between the points (2,-1) and (-1,-4). Round your answer to the nearest hundredth.
romanna [79]

Answer:

4.24

Step-by-step explanation:

8 0
3 years ago
A computer regularly priced at $843.75 is marked down 40%. What is the sale price of the computer. HELPP!!
yulyashka [42]

Answer: $506.25

Step-by-step explanation:

1. Find 40% of $843.75

- The decimal value of 40% is 0.4. We can multiply 0.4 by 843.75 to find the discount on the computer.

    -> 0.4*843.75= 337.5

2. Now that we have the discount, we can subtract 40% of the original price to find the new price.

843.75 - 337.5 = $506.25

8 0
3 years ago
Let z1 = 2 − 2i and z2 = (1 − i) + √3(1 + i).
gtnhenbr [62]

Answer:

Step-by-step explanation:

z₁ = 2 − 2i

z₂ = (1 − i) + √3(1 + i) = (1 + √3) + (√3 - 1) i

a) We get the modulus of z₁ as follows

║z₁║ = √((2)²+(-2)²) = 2

now we find the argument

α = Arctan (-2/2) = Arctan (-1) = -45º  ⇒   α = 360º + (-45º) = 315º

b) z₁ = 2 Cis 315º

Although the complex number is in binomic or polar form, its representation must be the same, since the complex number is the same, only that it is expressed in two different forms. The modulus represents the distance from the origin to the point. The degree of  rotation is the angle from the x-axis. When the polar form is expanded, the result is  the rectangular form of a complex number.

c) If  z₀*z₁ = z₂  and   z₀ = a + b i

we have

(a + b i)*(2 − 2i) = (1 + √3) + (√3 - 1) i

⇒  2a + 2bi - 2ai - 2bi² = (1 + √3) + (√3 - 1) i

⇒  2a + 2bi - 2ai - 2b(-1) = (1 + √3) + (√3 - 1) i

⇒  2a + 2b + 2bi - 2ai = (1 + √3) + (√3 - 1) i

⇒ 2 (a + b) + 2 (b - a) i = (1 + √3) + (√3 - 1) i

Now we can apply

2 (a + b) = 1 + √3

2 (b - a) = √3 - 1

Solving the system we get

a = 1/2

b = √3 / 2

Finally

z₀ = (1/2) + (√3 / 2) i

d) ║z₀║ = √((1/2)²+(√3 / 2)²) = 1

α = Arctan ((√3 / 2)/(1/2)) = 60º

e) z₀ = Cis 60º

f) Since z₂ = z₀*z₁, then z₂ is the transformation of z₁ rotated counterclockwise by arg(w) which is 60º

8 0
3 years ago
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