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STatiana [176]
3 years ago
5

How to write 1,900,000,000 in word form

Mathematics
1 answer:
-Dominant- [34]3 years ago
7 0
One billion nine hundred millions
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What's the relationship between the values of 4s in 44000
oksano4ka [1.4K]
The relationships of the 4's is 40,000 and 4,000
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2 years ago
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Consider a steel ball floating on the surface of mercury in a half-filled container. What happens when the rest of the container
Helen [10]

Answer

The steel ball will be caught up in between the mercury and water medium.

Step-by-step explanation:

  • Mercury with a density of 13546 kg/m^3
  • water with a density of 1000 kg/m^3
  • Steel with a density of 8050 kg/m^3

From the factual density of both substances given above ,it shows that mercury is more heavier than water. In that case mercury will be at the lower layer when mixed with water.

The steel ball will be caught in between both mixtures as the density of  steel is 8050 kg/m^3.Which  is more than Water but less than Mercury.

8 0
3 years ago
Without actual division prove that x4 +2x3 -2x2 +2x -3 is exactly divisible by<br> x2 +2x-3
Brut [27]

Answer:

Step-by-step explanation:                

\displaystyle\Large\boldsymbol{} x^4+2x^3-2x^2+2x-3= \\\\\\x^4+2x^2(x-1)+2x-\underbrace{2-1}_{-3}=  \\\\\\x^4-1+2x^2(x-1)+2x-2 = \\\\\\(x^2-1)(x^2+1)+2x^2(x-1)+2(x-1) = \\\\\\\underline{(x-1)}(x+1)(x^2+1)+2x^2\underline{(x-1)} +2\underline{(x-1)} =\\\\\\(x-1)((x+1)(x^2+1)+2x^2+2)=\\\\\\(x-1)(x^3+x^2+2x^2+x+2+1)=\\\\\\(x-1)(x^3+3x^2+x+3) =\\\\\\(x-1)( \underline{(x+3)}x^2+\underline{(x+3)})=(x^2+1)\underbrace{(x-1)(x+3)}_{x^2+2x-3}= \\\\\\(x^2+1)(x^2+2x-3) : (x^2+2x-3)=x^2+1

7 0
2 years ago
If one three-digit number ( 0 cannot be a left digit) is chosen at random from all those that can be made from the following set
Oksana_A [137]

The probability of choosing a number that is not a multiple of 2 is P = 0.44

<h3 /><h3>How to find the probability?</h3>

We need to count the number of options for each digit.

  • For the first digit, we have 8 options {1, 2, 3, 4, 5, 6, 7, 8}
  • For the second digit, we have 9 options {0 ,1, 2, 3, 4, 5, 6, 7, 8}
  • For the third digit, we have 9 options {0 ,1, 2, 3, 4, 5, 6, 7, 8}.

The total number of combinations is the product between the numbers of options:

C = 8*9*9 = 648

If we want our number to not be a multiple of 2 then it must end in a odd digit, the combinations that meet that condition are:

  • For the first digit, we have 8 options {1, 2, 3, 4, 5, 6, 7, 8}
  • For the second digit, we have 9 options {0 ,1, 2, 3, 4, 5, 6, 7, 8}
  • For the third digit, we have 4 options {1, 3,  5, 7}.

C = 8*9*4 = 288

Then the probability of selecting a 3 digit number that is not a multiple of 2 is:

P = 288/648 = 0.44

If you want to learn more about probability, you can read:

brainly.com/question/251701

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2 years ago
Combining like terms with negative coefficients
den301095 [7]

Answer:

you add them, but keep the sign

Step-by-step explanation:

8 0
3 years ago
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