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sattari [20]
2 years ago
7

A computer downloaded 20 kilobytes of data in 4 seconds. If it downloads data at a constant speed, it can download ______ kiloby

tes of data in 15 seconds
Mathematics
2 answers:
True [87]2 years ago
4 0

Answer:

75 kb

Step-by-step explanation:

Kaylis [27]2 years ago
3 0

Answer:75

Step-by-step explanation:

Divide 20 by 4 to get 5, the number of kb per second, and multiply by 15 to get 75

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Given that Roger is building a storage shed with wood blocks that are in the shape of cubic prisms.

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that means if width is 1 then height should be twice which is 2.


yes that is possible if we put one cubical prism over another cubical prism. then height of shed due to two prism will be twice than the width.

Hence correct choice should be "A. Yes. For every block of width, he could build two blocks high."

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He spent 9 dollars and 9 cents so $9.09
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The sum of 4 consecutive number is 330 what is the sum of the largest and the smallest number
Drupady [299]
X + x + 1 + x + 2 + x + 3 = 330
4x + 6 = 330
4x = 324, x = 81
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The answer is 165
3 0
1 year 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
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