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Lynna [10]
3 years ago
5

0.000000452 in scientific notation

Mathematics
2 answers:
iren2701 [21]3 years ago
6 0

0.000000452 in scientific notation would be 4.52 × 10^{-7}

Marta_Voda [28]3 years ago
5 0

Answer:

4.52 x 10^-7

Step-by-step explanation:

you have a very small number (numbers to right of decimal) so you're exponent will be negative.

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Volume= whl

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Multiply all (40), then multiply by the three times it was full.

Answer: 120ft^3

4 0
3 years ago
You get scores of 72 and 88 on 2 science tests. You have to have an average of at least 85 to not be grounded by your parents. W
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Answer:

95

Step-by-step explanation:

72+88+95=255:3= 85

7 0
3 years ago
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PLEASE HELP ME!! IM IN DESPERATE NEED OF HELP
vesna_86 [32]

Answer:

12

Step-by-step explanation:

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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
What value of z makes the statement true?
Shkiper50 [21]
-(z + (-17)) = 12 + 17

-(z - 17) = 12 + 17

Distribute -1:

-z + 17 = 12 + 17

Subtract 17 to both sides:

-z = 12

Multiply -1 to both sides:

z = -12
4 0
3 years ago
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