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kirza4 [7]
3 years ago
10

If f(x) = – x2 + 6x - 1 and g(x) = 3x - 1, find (f + g)(x).

Mathematics
1 answer:
lesantik [10]3 years ago
6 0

Answer:

-x^2 +9x -2

Step-by-step explanation:

f(x) = – x^2 + 6x - 1 and g(x) = 3x - 1,

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

               = – x^2 + 6x - 1 + 3x - 1

Combine like terms

              = -x^2 +9x -2

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2^x-2=16
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X=log(18)
————-
Log(2)

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3 0
3 years ago
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What are the x and y intercepts of the equation 2x-y=4
Len [333]

Answer:

x-intercept:(2,0)

y-intercept:(0,-4)

Step-by-step explanation:

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3 years ago
Write an equation in point-slope form, if m= 3 and passes through (4, 5).
Sergio039 [100]

Answer:

y - 5 = 3(x - 4)

General Formulas and Concepts:

<u>Algebra I</u>

Point-Slope Form: y - y₁ = m(x - x₁)  

  • x₁ - x coordinate
  • y₁ - y coordinate
  • m - slope

Step-by-step explanation:

<u>Step 1: Define</u>

Slope <em>m</em> = 3

Point (4, 5)

<u>Step 2: Write equation</u>

Point-Slope Form:   y - 5 = 3(x - 4)

5 0
3 years ago
How many complex zeros does the polynomial function have?<br> f(x)=−3x^6−2x^4+5x+6
REY [17]

one way would be to factor

I can't factor it so we will have to use Descartes' Rule of Signs which is helpful for finding how many real roots you have


it goes like this:

for a polynomial with real coefients, consider f(x)=-3x^6-2x^4+5x+6.

after arranging the terms in decending order in terms of degree, count how many times the signs of the coeffients change direction and minus 2 from that number until you get to 1 or 0. that will be the number of even roots the function can have

We have (-, -, +, +). the signs changed 1 times, so it has 1 real positive root


to get the negative roots, we evaluate f(-x) and see how many times the root changes

f(-x)=-3x^6-2x^4-5x+6

signs are (-, -, -, +). there was 1 change in sign

so the function has 1 real negative root



a total of 2 real roots


a function of degree n can have at most, n roots


our function is degree 6 so it has 6 roots

if 2 are real, then the others must be complex

6-2=4 so there are 4 complex roots


you can also show that there are only 2 real roots by using a graphing utility to see that there are only 2 real roots

7 0
3 years ago
Suppose that each of 935 smokers received a nicotine patch, which delivers nicotine to the bloodstream but at a much slower rate
nika2105 [10]

Answer:

26.2% of smokers,when given this treatment, would refrain from smoking for at least 6 months.

Step-by-step explanation:

In this problem, we have that:

935 smokers received the treatment(nicotine patch).

After 6 months, 245 of them were not smoking.

Assuming it is reasonable to regard this sample as representative of all smokers, estimate the percentage of all smokers who, when given this treatment, would refrain from smoking for at least 6 months.

This percentage is 245/935 = 0.2620.

26% of smokers,when given this treatment, would refrain from smoking for at least 6 months.

7 0
3 years ago
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