All categories

3 years ago

What is the multiple zero and multiplicity of f(x)=x^3+2x^2+x? Help needed !!

Mathematics

2 answers:

3241004551 [841]3 years ago

7 0

f(x) = x³ + 2x² + x

f(x) = x.(x² + 2x + 1)

f(x) = x.(x + 1)²

So, f(x) = 0 when x = 0 or when x = -1

But

f(x) = x.(x + 1)² = x.(x + 1).(x + 1)

So we have three roots, x = 0, x = -1 and x = -1, although two of them are equal.

So we have a multiple zero x = -1 with a multiplicity 2.

Alterativa A.

tatiyna3 years ago

7 0

Answer:

Step-by-step explanation:

You might be interested in

A culture of bacteria has an initial population of 22000 bacteria and doubles every 5 hours. Using the formula P t = P 0 ⋅ 2 t d

Bas_tet [7]

We have been given that a culture of bacteria has an initial population of 22,000 bacteria and doubles every 5 hours. Using the formula $P(t) = P_0\cdot 2^{\frac{t}{d}}$ , where P(t) is the population after t hours, $P_0$ is the initial population, t is the time in hours and d is the doubling time.

We are asked to find the population of bacteria after 17 hours.

First of all, we will substitute our given values in doubling life formula as:

$P(t) =22,000\cdot 2^{\frac{t}{5}}$

Now to find population of bacteria after 17 hours, we will substitute $t=17$ in our formula as:

$P(117) =22,000\cdot 2^{\frac{17}{5}}$

$P(117) =22,000\cdot 2^{3.4}$

$P(117) =22,000\cdot (10.556063286183)$

$P(17) =232233.392296026$

$P(17)\approx 232,233$

Therefore, the bacteria population will be 232,233 after 17 hours.

7 0

4 years ago

Is the relationship shown by the data linear? If so, model the data with an equation.

zhenek [66]

Answer:

Therefore the equation of the line through ( -7 , 5 ) and ( -5 , 9) is

Linear Relationship i.e $y-5=2(x+7)$

Step-by-step explanation:

Given:

Let,

point A( x₁ , y₁) ≡ ( -7, 5 )

point B( x₂ , y₂) ≡ (-5, 9)

To Find:

Equation of Line AB =?

Solution:

Equation of a line passing through Two points A( x₁ , y₁) and B( x₂ , y₂)is given by the formula

$(y - y_{1} )=(\frac{y_{2}-y_{1} }{x_{2}-x_{1} })\times(x-x_{1}) \\$

Substituting the given values in a above equation we get

$(y-(5))=(\frac{9-(5)}{-5--7})\times (x--7)\\ \\(y-5)=\frac{4}{2}(x+7)\\\\y-5=2(x+7)...............\textrm{which is the required equation of the line AB}$