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krek1111 [17]
3 years ago
9

I WILL GIVE BRAINLY IF UR RIGHT

Mathematics
2 answers:
romanna [79]3 years ago
5 0
The last one (the 64c one)
Y_Kistochka [10]3 years ago
5 0

Answer:

you realize that im just getting more free points right

Step-by-step explanation:

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How do i solve 426 to nearest ten
puteri [66]

We look at the tens place

20 is in the tens place

now look at the Ones place

6 is in the Ones place

If it is 5 or greater we round it to the next ten, if it is 4 or less we keep it the same

6 is greater than 5

The number will be rounded to 430

5 0
3 years ago
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What is <br> 12 m/mi = cm/s
yarga [219]

<em>1 Cancel m.</em>

12i=c*m/s

<em>2 Use rule a*b/c = ab/c</em>

12i=cm/s

<em>3 Multiply both sides by s.</em>

12is=cm

<em>4 Divide both sides by 12.</em>

is=cm/12

<em>5 Divide both sides by i</em>

s=cm/12/i

<em>6 Simplify cm/12/i</em>

s=sm/12i

4 0
2 years ago
. What is the GCF of the expression: 28xy-14X
Tamiku [17]

Answer:

14 x

Step-by-step explanation:

3 0
2 years ago
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
<img src="https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%264%5C%5C8%26-5%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%
brilliants [131]

Answer:

What's this? I don't understand.

4 0
3 years ago
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