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rosijanka [135]
3 years ago
15

I have 6 digits. One of my 3s is worth 300,000. The other is worth 1/10 as much. My 6 is worth 600. The rest of my digits are ze

ros.
What number am I?
Mathematics
2 answers:
slavikrds [6]3 years ago
8 0

Answer:

330600

Step-by-step explanation:

Given that there are six digits in a number

Out of this 6 digits, one digit has value as 300000

This is possible only if I digit is 3 since place value here is only 100000

\frac{1}{10} (300000) = 30000 implies that the second digit in place value of 10000 is also 3.

When 6 has value as 600, then 6 is in 100th place.  Other digits are 0

Hence the number is

330600

Anna35 [415]3 years ago
4 0
The answer is:                         
330,600.
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Lina20 [59]

Answer:

(4x+5y=7

y=3x+9

4x+5(3x+9)=7

4x+15x+45=7

19x=7-45

19x= -38

19×/19=-38/19

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whlie y=3x+9

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y= -6+9

y=3 end solution

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Which two of the following are characteristics of surveys?
Hatshy [7]

Answer:

<em>B. Statistical analysis is applied to the results of the study.</em>

<em>B</em><em> </em><em>and </em><em>C.</em>

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At 3pm, the temperature outside was 5 1/5 degrees Fahrenheit. The temperature then fell steadily by 2 1/2 degrees per hour for t
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Y = f(x) has the derivative f'(x) = (x + 1)²(x + 3)(x² + 2mx + 5) with ∀x∈i.
maksim [4K]

9514 1404 393

Answer:

  m ≥ -√5

Step-by-step explanation:

If g(x) = f(|x|) for x∈i, then g(x) = f(x) for x∈ℝ: x ≥ 0.

f'(x) is 5th-degree, so f(x) is 6th-degree, meaning it is generally U-shaped. Since we're only concerned with x ≥ 0, we want to make sure f'(x) has no real zeros of odd multiplicity such that x > 0. The given factors of f'(x) make it have real zeros at x = -3 and x = -1.

For the last factor, (x² +2mx +5) to have no positive real zeros of odd multiplicity, we must have m ≥ 0 or the discriminant ≤ 0. The discriminant is ...

  d = b² -4ac = (2m)² -4(1)(5) = 4m² -20 . . . . . discriminant of the last factor

  d ≤ 0 . . . . . . . . . . the condition for no real zeros

  4m² -20 ≤ 0

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  m² ≤ 5 . . . . . . . . .add 5

  |m| ≤ √5 . . . . . . . take the square root

This tells us there will be a positive real zero of multiplicity 2 in f'(x) when m = -√5, and there will be no positive real zeros for -√5 < m < 0

There will be no odd-multiplicity positive real zeros in the derivative function f'(x) as long as m ≥ -√5. This means the slope of f(x) is non-negative for x ≥ 0, hence f(|x|) has its only minimum at x=0.

_____

<em>Additional comment</em>

The multiplicity of the zeros of f'(x) is important because the derivative will only change sign where the multiplicity is odd. When the discriminant of (x²+2mx+5) is zero, the associated positive real zero will have multiplicity 2, hence f'(x) will not change sign there.

3 0
2 years ago
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