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Brums [2.3K]
3 years ago
15

PLEASE HELP ASAP Determine the period.

Mathematics
1 answer:
kolbaska11 [484]3 years ago
6 0
That’s easy_________
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Solve the following equation:
stealth61 [152]

Answer: Y=1/8

Slope m=0, b =1/8

Step-by-step explanation:

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2 years ago
Point M lies between points L and N on .
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Answer: The answer is d. 64 units

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Graph g(x), where f(x) = 2x − 5 and g(x) = f(x + 1).
pav-90 [236]

Answer:

B.) a line labeled g(x) that passes through points 0, negative 3 and 4, 5

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I graphed both equation on the graph below to find the description of the graph of g(x).

5 0
3 years ago
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Find the real numbers x and y if -3+ix^2y and x^2+y+4i are conjugate of each other. Pls solve with the steps
Firdavs [7]
ANSWER
x = ±1 and y = -4.
Either x = +1 or x = -1 will work

EXPLANATION
If -3 + ix²y and x² + y + 4i are complex conjugates, then one of them can be written in the form a + bi and the other in the form a - bi. In other words, between conjugates, the imaginary parts are same in absolute value but different in sign (b and -b). The real parts are the same

For -3 + ix²y
⇒ real part: -3
⇒ imaginary part: x²y

For x² + y + 4i
⇒ real part: x² + y (since x, y are real numbers)
⇒ imaginary part: 4

Therefore, for the two expressions to be conjugates, we must satisfy the two conditions. 

Condition 1: Imaginary parts are same in absolute value but different in sign. We can set the imaginary part of -3 + ix²y to be the negative imaginary part of x² + y + 4i so that the 

   x²y = -4 ... (I)

Condition 2: Real parts are the same

   x² + y = -3 ... (II)

We have a system of equations since both conditions must be satisfied

   x²y = -4 ... (I)
   x² + y = -3 ... (II)

We can rearrange equation (II) so that we have

   y = -3 - x² ... (II)

Substituting into equation (I)

   x²y = -4 ... (I)
   x²(-3 - x²) = -4
   -3x² - x⁴ = -4
   x⁴ + 3x² - 4 = 0
   (x² + 4)(x² - 1) = 0
   (x² + 4)(x-1)(x+1) = 0

Therefore, x = ±1.
Leave alone (x² + 4) as it gives no real solutions.

Solve for y:

   y = -3 - x² ... (II)
   y = -3 - (±1)²
   y = -3 - 1
   y = -4

So x = ±1 and y = -4. We can confirm this results in conjugates by substituting into the expressions:

   -3 + ix²y 
   = -3 + i(±1)²(-4)
   = -3 - 4i

   x² + y + 4i
   = (±1)² - 4 + 4i
   = 1 - 4 + 4i
   = -3 + 4i

They result in conjugates
4 0
3 years ago
Read 2 more answers
Consider the first four terms of an arithmetic sequence below.
VMariaS [17]
7, 13, 19 and 25 have a common difference:  6.  
6 added to 7 gives us 13; 6 added to 13 gives us 19, and so on.

Explicit formula:  a(n) = 7 + 6(n-1), where 7 is the first term and n is the counter (1, 2, 3, ...).

The first term is 7 (given).  This corresponds to n=1.

The second term is a(2) = 7 + 6(2-1), or 7 + 6, or 13.  This corresponds to n = 2.

and so on.


6 0
3 years ago
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