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lakkis [162]
3 years ago
6

PLease anwer the following ( > o < )

Mathematics
1 answer:
mina [271]3 years ago
4 0

Answer: 12 = x + 4

Step-by-step explanation:

We know that the length of the green line is 12.

And the length of the black line will be equal to the sum of the lengths of the two smaller segments, x, and 4

Then the length of the black line is x + 4

We want to write an equation that says that the length of the green line is equal to the length of the black line, this is:

Length of green line = Length of black line

12 = x + 4

That is the equation we wanted.

Now we could solve it if we subtract 4 in both sides of the equation:

12 - 4 = (x + 4) - 4

8 = x

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In mrs kuliks class there are 20 female students and 8 male students what is the percent of male students
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Answer:

About 29%

Step-by-step explanation:

8 0
3 years ago
What is a decimal that is equivalent to the expression (3 x 10) + (4•1) + ( 7 • 1/10) + ( 5• 1/100)
riadik2000 [5.3K]

The decimal form of the answer to that expression is

34.75

6 0
3 years ago
Brandon earns a rebate each month for his meals and movie tickets he earns $.50 for each movie ticket and 2 dollars for each mea
bezimeni [28]
$25=0.50(x)+2(y)
$25=0.50(22)+2(y)
8 0
3 years ago
Section 5.2 Problem 6:<br><br>Find the general solution<br><img src="https://tex.z-dn.net/?f=y%27%27%20%2B%206y%27%20%2B%2010y%2
mihalych1998 [28]

Answer:

y=e^{-3t}(A\: cos\: t+B\:sin\:t)

Step-by-step explanation:

<u>Given Second-Order Homogenous Differential Equation</u>

y''+6y'+10y=0

<u>Use Auxiliary Equation</u>

<u />m^2+6m+10=0\\\\(m+3)^2+1=0\\\\(m+3)^2=-1\\\\m+3=\pm i\\\\m=-3\pm i

<u>General Solution</u>

<u />y=e^{pt}(A\: cos\: qt+B\:sin\:qt)\\\\y=e^{-3t}(A\: cos\: t+B\:sin\:t)

Note that the DE has two distinct complex solutions p\pm qi where A and B are arbitrary constants.

4 0
2 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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