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love history [14]
3 years ago
11

6x^2-5x+3 - (3x^2+8x-4)

Mathematics
2 answers:
amid [387]3 years ago
5 0
The answer should be 3x^2 -13x+7.
pantera1 [17]3 years ago
3 0

6x {}^{2}  - 5x + 3 - ( {3x}^{2}  + 8x - 4)

= 6 {x}^{2}  - 5x + 3 -  {3x}^{2}  - 8x + 4

= ( {6x}^{2}  -  {3x}^{2} ) + ( - 5x - 8x) + (3 + 4)

=  {3x}^{2}  - 13x + 7

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I need help??????????????
Likurg_2 [28]

Answer:

20

Step-by-step explanation:

50*0.40=20

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2 years ago
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Correct answer gets brainliest.
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The answer is C based on SAS
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A bag contains three red marbles, two green ones, one lavender one, four yellows, and five orange marbles. HINT [See Example 7.
kirill115 [55]

Answer:

There are 12 sets of four marbles include all the red ones.    

Step-by-step explanation:

Given : A bag contains three red marbles, two green ones, one lavender one, four yellows, and five orange marbles.

To find : How many sets of four marbles include all the red ones?

Solution :

Number of red marbles = 3

Number of green marbles = 2

Number of lavender marbles = 1

Number of yellow marbles = 4

Number of orange marbles = 5

We have to form sets of four marbles include all the red ones,

For position of getting red ones we have three red marbles i.e. ^3C_3

For the fourth one we have 12 choices i.e. ^{12}C_1

Total sets of four marbles include all the red ones is

=^3C_3\times ^{12}C_1

=1\times 12

=12

Therefore, There are 12 sets of four marbles include all the red ones.

5 0
3 years ago
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
elena-s [515]

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
Mr Jacobson asked her student to write a word problem that describes the graph below
bazaltina [42]

Answer:

I'm gonna need to see the graph, sweetheart.


8 0
3 years ago
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