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sesenic [268]
3 years ago
15

Ok I have a couple of questions please help on the one you know please!!

Mathematics
1 answer:
m_a_m_a [10]3 years ago
6 0

Answer to Roberto's question:

4

Step-by-step explanation:

I'm not sure how to explain this one (my apologies).

Answer to Parallelogram JKLM question:

Q = 110° & ST = 9 ft

Explanation:

When reflecting this, the angle/side lengths remain the same. That means the information is going to be the exact same. All you have to do is figure out what corresponds to what from JKLM to QTSR.

If we reflect a figure across a horizontal reflection line through its center, the figure will just flip sides. Making L = K, M = J, and so on.

Basically, flip the shape upside down for this problem, and look at that and the shape without angles, find the corresponding letter to the angle to get your answer. In this case, Q is that corresponding letter. Meaning your answer for Q = 110°

When it comes to figuring out ST, since the shape is just flipped and all the lines making the parallelogram are equal to each other, the feet remains the same. Meaning ST = 9 ft.

Answer for a triangle with vertices equation:

(-6,7) (2,8) (-4,3)

This one I didn't know how to do but I looked up the problem and this was the answer.

Answer to Hailey's plywood question:

7 inches

Explanation:

This one's a trick question. The size of an object does not change no matter how much you rotate it. Therefore the answer is going to be the same.

Sorry this took so long, hope I helped!

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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
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