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creativ13 [48]
2 years ago
9

According to the Fundamental Theorem of Algebra, the graph of f(x) = x2 - 4x + 3, has roots. From the graph we can see that it h

as zeros.

Mathematics
1 answer:
andreyandreev [35.5K]2 years ago
3 0

Answer:

The graph f(x)=x^2-4x+3  has two zeros namely 3 and 1.

Step-by-step explanation:

Consider the given equation of graph f(x)=x^2-4x+3

According to the Fundamental Theorem of Algebra

For a given polynomial of degree n can have a maximum of n roots.

Thus, for the given equation f(x)=x^2-4x+3  the degree of polynomial is 2 , thus the function can have maximum of 2 roots.

We know at roots the value of function is 0 that is f(x) = 0,

Substitute f(x) = 0 , we get, f(x)=x^2-4x+3=0

This is a quadratic equation, x^2-4x+3=0

We first solve it manually and then check by plotting graph.

Quadratic equation can be solved using middle term splitting method,

here, -4x can be written as -x-3x,

x^2-4x+3=0 \Rightarrow x^2-x-3x+3=0

\Rightarrow x(x-1)-3(x-1)=0

\Rightarrow (x-3)(x-1)=0

Using zero product property, a\cdot b=0 \Rightarrow a=0\ or \ b=0

\Rightarrow (x-3)=0 or \Rightarrow (x-1)=0

\Rightarrow x=3 or \Rightarrow x=1

Thus, the two zero of f(x) are 3 and 1.

We can also see on graph attached below that the graph f(x)=x^2-4x+3  has two zeros  namely 3 and 1.  

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A species of beetles grows 32% every year. Suppose 100 beetles are released into a field. How many beetles will there be in 10 y
siniylev [52]

Given that a species of beetles grows 32% every year.

So growth rate is given by

r=32%= 0.32


Given that 100 beetles are released into a field.

So that means initial number of beetles P=100


Now we have to find about how many beetles will there be in 10 years.

To find that we need to setup growth formula which is given by

A=P(1+r)^n where A is number of beetles at any year n.

Plug the given values into above formula we get:

A=100(1+0.32)^n

A=100(1.32)^n


now plug n=10 years

A=100(1.32)^{10}=100(16.0597696605)=1605.97696605

Hence answer is approx 1606 beetles will be there after 20 years.


Now we have to find about how many beetles will there be in 20 years.

To find that we plug n=20 years

A=100(1.32)^{20}=100(257.916201549)=25791.6201549

Hence answer is approx 25791 beetles will be there after 20 years.



Now we have to find time for 100000 beetles so plug A=100000

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100000=100(1.32)^n

1000=(1.32)^n

log(1000)=n*log(1.32)

\frac{\log\left(10000\right)}{\log\left(1.32\right)}=n

33.174666862=n

Hence answer is approx 33 years.

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