3 years ago

What are the zeros of the function? what are their multiplicities? f(x)=x^4-4x^3+3x^2

1 answer:

4 0

Since each term contains x, x= 0 is one answer Also as x^2 is a factor it has duplicity 2.
Factoring:-

x^2 (x^2 - 4x + 3) = 0
x^2( x - 3)(x - 1) = 0

so x = 3 and x = 1 are also zeros

Answer:- x = 0 (duplicity 2), x = 1 and x = 3.

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Step-by-step explanation:

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3 years ago

the combined area of two squares is 45 square cm. each side one square is twice as long as a side of the other Square. what is t

anygoal [31]

6 centimeters

x will represent the side length of the smaller square, so 2x is the side length of the bigger square.

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Please consider marking this answer as Brainliest to help me advance.

4 0

3 years ago

Solve for y. y - 12 = -10 y = 2 y = -22 y = 22 y = -2

ira [324]

Answer:

y - 12 = -10

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is the answer

3 0

3 years ago

Read 2 more answers

Find r(t) if r'(t) = 5t4i + 6t5j + t k and r(1) = i + j.

Vladimir [108]

$\mathbf r(t)=\displaystyle\int(5t^4\,\mathbf i+6t^5\,\mathbf j+t\,\mathbf k)\,\mathrm dt$
$\mathbf r(t)=(t^5+C_1)\,\mathbf i+(t^6+C_2)\,\mathbf j+\left(\dfrac12t^2+C_3\right)\,\mathbf k$

Given $\mathbf r(1)=\mathbf i+\mathbf j$ , we have

$\begin{cases}1^5+C_1=1\\\\1^6+C_2=1\\\\\dfrac121^2+C_3=0\end{cases}\implies C_1=0,C_2=0,C_3=-\dfrac12$

so that the particular solution is

$\mathbf r(t)=t^5\,\mathbf i+t^6\,\mathbf j+\dfrac12(t^2-1)\,\mathbf k$

8 0

3 years ago

Can someone help quick

9966 [12]

Step-by-step explanation:

22+2x+8= x+20

2x+30=x+20

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answer is -10

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3 years ago