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Andru [333]
2 years ago
10

9. Write a polynomial function in factored form with zeros at -2, 5, and 6.

Mathematics
1 answer:
Amanda [17]2 years ago
8 0

Answer:

f(x) = ( x+2)(x-5)(x-6)

Step-by-step explanation:

We can write the equation with zeros at b1,b2,b3

f(x) =a( x-b1)(x-b2)(x-b3)  where a is a constant and b1 b2 b3 are the zeros

f(x) = a( x- -2)(x-5)(x-6)

f(x) = a( x+2)(x-5)(x-6)

We can choose a since we are not given any more information about the function.  Let a = 1

f(x) = 1( x+2)(x-5)(x-6)

f(x) = ( x+2)(x-5)(x-6)

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Simplify the equation
Arturiano [62]

Answer:

\dfrac{16x^3-37x^2-16x+37}{x^2+2}

Step-by-step explanation:

There aren't any factors that cancel (except 3x). The best you can do is multiply it out.

\displaystyle\frac{\frac{3x^2-3}{x}}{\frac{3(x^2+2)}{16x^2-37x}}=\frac{3(x^2-1)}{x}\cdot\frac{16x^2-37x}{3(x^2+2)}\\\\=\frac{x(x^2-1)(16x-37)}{x(x^2+2)}=\frac{16x^3-37x^2-16x+37}{x^2+2}

7 0
3 years ago
What is the sum of the first 50 terms of the series 2 + 17 + 32 + 47+ ...?​
Alex

Answer:

\huge\boxed{18,475}

Step-by-step explanation:

The terms of this sum make the arithmetic sequence.

The fomula of a sum of <em>n</em> terms of an arithmetic sequence:

S_n=\dfrac{[2a_1+(n-1)d]\cdot n}{2}

We have

a_1=2\\d=17-2=32-12=47-32=15\\n=50

Substitute:

S_{50}=\dfrac{[2\cdot2+(50-1)\cdot15]\cdot50}{2}=(4+49\cdot15)\cdot25=(4+735)\cdot25\\\\=739\cdot25=18,475

3 0
3 years ago
Find f ∘ g f∘g and g ∘ f g∘f. f ( x ) = x 2 / 3 , g ( x ) = x 9 f(x)=x2/3, g(x)=x9 (a) f ∘ g f∘g (b) g ∘ f g∘f Find the domain o
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How much of a radioactive kind of thorium will be left after 14,680 years if you start with
babymother [125]

Answer:

8978 grams

Step-by-step explanation:

The equation to find the half-life is:

N(t)= N_{0}e^{-kt}

N(t) = amount after the time <em>t</em>

N_{0} = initial amount of substance

t = time

It is known that after a half-life there will be twice less of a substance than what it intially was. So, we can get a simplified equation that looks like this, in terms of half-lives.

N(t)= N_{0}e^{-\frac{ln(\frac{1}{2}) }{t_{h} } t} or more simply N(t)= N_{0}(\frac{1}{2})^{\frac{1}{t_{h} } }

t_{h} = time of the half-life

We know that N_{0} = 35,912, t = 14,680, and t_{h}=7,340

Plug these into the equation:

N(t) = 35912(\frac{1}{2})^{\frac{14680}{7340} }

Using a calculator we get:

N(t) = 8978

Therefore, after 14,680 years 8,978 grams of thorium will be left.

Hope this helps!! Ask questions if you need!!

8 0
2 years ago
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