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Marianna [84]
3 years ago
5

If f(x)=2x-5 and g(x)=x^2-4x-8, find (f+g)(x).

Mathematics
2 answers:
viktelen [127]3 years ago
8 0

f(x)=2x-5,\ g(x)=x^2-4x-8\\\\(f+g)(x)=f(x)+g(x)=(2x-5)+(x^2-4x-8)\\\\=2x-5+x^2-4x-8=x^2+(2x-4x)+(-5-8)=x^2-2x-13

snow_tiger [21]3 years ago
3 0

Answer:

\text{The value of (f+g)(x) is }x^2-2x-13

Step-by-step explanation:

Given the two functions

f(x)=2x-5

g(x)=x^2-4x-8

we have to find the value of (f+g)(x)

As we know

(f+g)(x)=f(x)+g(x)=(2x-5)+(x^2-4x-8)

=2x-5+x^2-4x-8

=x^2+2x-4x-5-8=x^2-2x-13

\text{Hence, the value of (f+g)(x) is }x^2-2x-13

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After a discount of 40%, the sale price of a ride pass at a fun park is $30. What is the original
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Answer:

y = 3/4x + 19/2

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m = 3/4. point (-6,5)

Point slope form:

y-5=3/4(x- -6)

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Consider the game of independently throwing three fair six-sided dice. There are six combi- nations in which the three resulting
Murrr4er [49]

Answer:

See explanation below.

Step-by-step explanation:

1) First let's take a look at the combinations that sum up 10:

  1. 1+3+ 6,
  2. 1+ 4+ 5,
  3. 2+2+6,
  4. 2+3+5,
  5. 2 + 4 + 4,
  6. 3+3+4

Notice that when we have 3 different numbers on the dice, we can permute them in 6 different ways. For example: Let's take 1 + 3 + 6, we can get this sum with these permutations:

1 + 3 + 6, 1 + 6 + 3, 3 + 6 + 1, 3 + 1 + 6, 6 + 1 + 3, 6 + 3 + 1.

And when we have two different numbers on the dice, we can permute them in 3 different ways:

2 + 2 + 6, 2 + 6 +2, 6 + 2 + 2.

So now we're going to write down the 6 combinations that sum up 10 but we're going to write down how many permutations of them we get:

  1. 1+3+ 6 : 6 permutations
  2. 1+ 4+ 5 : 6 permutations
  3. 2+2+6: 3 permutations
  4. 2+3+5: 6 permutations
  5. 2 + 4 + 4: 3 permutations
  6. 3+3+4: 3 permutations

Total of permutations: 6 + 6 + 3 + 6 + 3 + 3 =27.

Thus we have 27 different ways of getting a sum of 10.

2) Now we're going to take a look at the combinations that sum up 9 and we're going to proceed in a similar way:

  1. 1 + 2 + 6: 6 permutations
  2. 1+3+5: 6 permutations
  3. 1+4+4: 3 permutations
  4. 2+ 3+ 4: 6 permutations
  5. 2+2 +5: 3 permutations
  6. 3+3+3: 1 permutation.

Total of permutations: 6 + 6 + 3 + 6 +3 + 1 = 25.

Thus we have 25 different ways of getting a sum of 10

And we can conclude that the probability of getting a total of 10 is larger than the probability to get a total of 9.

5 0
4 years ago
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