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Sliva [168]
3 years ago
10

Find the value of k if it is known that the graph of y=kx+2 goes through the point. a(14,-40)

Mathematics
2 answers:
Alex_Xolod [135]3 years ago
8 0

Answer:

k=-3

Explanation:

Plug in x and y values

-40=k(14)+2

-2 both sidea

-42=k(14)

divide by 14 both sides

k=-3

zimovet [89]3 years ago
8 0

k = -3

Step-by-step explanation:

y = kx + 2 through the point a(14, -40)

Subtitute…

y = kx + 2

-40 = k(14) + 2

-40 = 14k + 2

-40 - 2 = 14k

-42 = 14k

(-42)/14 = k

<u>-3 = k</u>

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There are two answers to this question
the first possible answer is 5, 6, and 7 and the second possible answer is -4, -3, and -2


Step-by-step explanation:

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the product of the smallest and largest number is 17 more than 3 times the middle number, or
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Step-by-step explanation:

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a sample of 100 pregnancies were found to take an average of 267 days. the population deviation is known to be 13 days. find a 9
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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

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