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Sunny_sXe [5.5K]
3 years ago
11

If f(x)=2x-1 and g(x)=x^2-3x-2 find (f+g)(x)

Mathematics
2 answers:
Pachacha [2.7K]3 years ago
4 0

Answer:

(f+g)(x) =  x²-x-3

Step-by-step explanation:

We have given functions.

f(x)=2x-1 and g(x)=x²-3x-2

We have to find the sum of two functions.

(f+g)(x) = ?

The formula to find the sum of two function:

(f+g)(x) = f(x) + g(x)

Putting given values in above formula, we have

(f+g)(x) = 2x-1+x²-3x-2

Adding like terms, we have

(f+g)(x) = x²+(-3+2)x+(-1-2)

(f+g)(x) =  x²+(-1)x+(-3)

(f+g)(x) =  x²-x-3 which is the answer.

Komok [63]3 years ago
3 0

Answer:

(f+g)(x)=x^2-x-3.

Step-by-step explanation:

The given functions are;

f(x)=2x-1

and

g(x)=x^2-3x-2

Recall that;

(f+g)(x)=f(x)+g(x)

\Rightarrow (f+g)(x)=2x-1+x^2-3x-2.

Regroup the like terms to get;

(f+g)(x)=x^2-3x+2x-2-1.

Simplify;

(f+g)(x)=x^2-x-3.

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Bad White [126]

Answer:

A.

Step-by-step explanation:

Since it says greater than, that means that it cannot be B or D. And ≥ means greater than or equal to and they don't state that in the problem. So the answer would be A.

I HOPE THIS HELPED! :)

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The accompanying data came from a study of collusion in bidding within the construction industry. No. Bidders No. Contracts 2 6
viktelen [127]

Answer:

a)

The proportion of contracts involved at most five bidders is 0.667.

The proportion of contracts involved at least five bidders is 0.51.

b)

The  proportion of contracts involved  between five and 10 inclusive bidders is 0.5.

The  proportion of contracts involved  strictly between five and 10 bidders is 0.304.

Step-by-step explanation:

No. bidders    No. contracts     Relative frequency of contracts

2                         6                       6/102=0.0588

3                         20                     20/102=0.1961

4                         24                     24/102=0.2353

5                         18                       18/102=0.1765

6                         13                        13/102=0.1275

7                          7                          7/102=0.0686

8                          5                          5/102=0.049

9                          6                          6/102=0.0588

10                         2                           2/102=0.0196

11                          1                            1/102=0.0098

Total                  102

a)

We have to find proportion of contracts involved at most five bidders.

Proportion of at most 5= Relative frequency 2+ Relative frequency 3+ Relative frequency 4+ Relative frequency 5

Proportion of at most 5=0.0588+ 0.1961+0.2353+0.1765

Proportion of at most 5=0.6667

The proportion of contracts involved at most five bidders is 0.667.

proportion of at least five bidders= proportion≥5= 1- proportion less than 5

Proportion less than 5=0.0588+ 0.1961+0.2353=0.4902

proportion of at least five bidders=1-0.4902=0.5098

The proportion of contracts involved at least five bidders is 0.51

b)

We have to find proportion of contracts involved  between five and 10 inclusive bidders.

Proportion of contracts between five and 10 inclusive= Relative frequency 5+ Relative frequency 6+ Relative frequency 7+ Relative frequency 8+ Relative frequency 9+ Relative frequency 10

Proportion of between five and 10 inclusive=0.1765 +0.1275 +0.0686 +0.049 +0.0588 +0.0196

Proportion of between five and 10 inclusive=0.5

The  proportion of contracts involved  between five and 10 inclusive bidders is 0.5

We have to find proportion of contracts involved  strictly between five and 10 bidders.

Proportion of contracts strictly between five and 10=  Relative frequency 6+ Relative frequency 7+ Relative frequency 8+ Relative frequency 9

Proportion of strictly between five and 10=0.1275 +0.0686 +0.049 +0.0588

Proportion of strictly between five and 10=0.3039

The  proportion of contracts involved  strictly between five and 10 bidders is 0.304.

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Assoli18 [71]

Answer:

1. 54ft squared  2. 48 cm squared

Step-by-step explanation:

6 0
3 years ago
Helpppppp pleaseeeeeee
NeX [460]

Answer: b)

Step-by-step explanation:

You can rule out A) and C), because you have to find area (no adding.)

To do this problem, you can individually multiply the estimated hieght and

width by 3 and then multiply together.

This is because there are 3 walls that each are 9 3/4 tall and 14 1/4 wide. To find total height and width, multiply by 3. Then find total area (lxw)

9 3/4 rounds to 10

14 1/4 rounds to 14

= 3 (10) x 3 (14)

D) is incorrect because you have to multiply each number by 3. (Or you can multiply by 6)

It is B)

4 0
3 years ago
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