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bagirrra123 [75]
3 years ago
6

Find the distance between the two points (0,0), (4,3)

Mathematics
1 answer:
Amiraneli [1.4K]3 years ago
4 0

Answer:

Distance between the points=5

Step-by-step explanation:

The distance between two points in coordinate geometry is measured by using the distance formula:

If we have given two points

                   (x_1,y_1)\\\\(x_2,y_2)

Distance formula=  \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

For (0,0) and (4,3)

      Distance:                          

                        \sqrt{(4-0)^2+(3-0)^2} \\\\\sqrt{4^2+3^2}\\\\\sqrt{16+9}\\\\\sqrt{25}

Or

      Distance between the points=5

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For f(x) = 4x+1 and g(x)=x²-5, find (f-g)(x).
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Answer: -x² + 4x + 6

Given:

f(x) = 4x + 1

g(x) = x² - 5

Solve:

= (f - g)(x)

= f(x) - g(x)

= 4x + 1 - (x² - 5)

= 4x + 1 - x² + 5

= -x² + 4x + 6

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Set the two equal to each other and solve for x

-2x+1 = -2x²+1
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Joe has 4/5 as much money as bob. Tim has $50 more than joe. Rita has $150, which is $35 more than bob. How much money does tim
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Rita = $150

Bob = $150 - $35 = $115    Bob has $115

Joe = $115 divided by 5 = $23      $23 x 4 = $92    Joe Has $92

Tim = Has $50 more then joe, $92 + $50 = $142     Tim has $142

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Four cards are dealt from a standard fifty-two-card poker deck. What is the probability that all four are aces given that at lea
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Answer:

The probability is 0.0052

Step-by-step explanation:

Let's call A the event that the four cards are aces, B the event that at least three are aces. So, the probability P(A/B) that all four are aces given that at least three are aces is calculated as:

P(A/B) =  P(A∩B)/P(B)

The probability P(B) that at least three are aces is the sum of the following probabilities:

  • The four card are aces: This is one hand from the 270,725 differents sets of four cards, so the probability is 1/270,725
  • There are exactly 3 aces: we need to calculated how many hands have exactly 3 aces, so we are going to calculate de number of combinations or ways in which we can select k elements from a group of n elements. This can be calculated as:

nCk=\frac{n!}{k!(n-k)!}

So, the number of ways to select exactly 3 aces is:

4C3*48C1=\frac{4!}{3!(4-3)!}*\frac{48!}{1!(48-1)!}=192

Because we are going to select 3 aces from the 4 in the poker deck and we are going to select 1 card from the 48 that aren't aces. So the probability in this case is 192/270,725

Then, the probability P(B) that at least three are aces is:

P(B)=\frac{1}{270,725} +\frac{192}{270,725} =\frac{193}{270,725}

On the other hand the probability P(A∩B) that the four cards are aces and at least three are aces is equal to the probability that the four card are aces, so:

P(A∩B) = 1/270,725

Finally, the probability P(A/B) that all four are aces given that at least three are aces is:

P=\frac{1/270,725}{193/270,725} =\frac{1}{193}=0.0052

5 0
3 years ago
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