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3 years ago

polynomial function of degree 4 with -1 as a zero of multiplicity 3 and 0 as a zero of multiplicity 1

1 answer:

6 0

$\bf \begin{cases}
x=-1\implies &x+1=0\\
x=0\implies &x=0
\end{cases}
\\\\\\
\stackrel{\textit{multiplicity of 3}}{(x+1)^3}~~\stackrel{\textit{multiplicity of 1}}{(x)^1}=\stackrel{original~polynomial}{0}
\\\\\\
(x^3+3x^2+3x+1)(x)=\stackrel{y}{0}\implies x^4+3x^3+3x^2+x=y$

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Doug hits a baseball straight towards a 15 ft high fence that is 400 ft from home plate. The ball is hit 2.5ft above the ground

Scorpion4ik [409]

Answer:

$125.4\ \text{m/s}$

Step-by-step explanation:

u = Initial velocity of baseball

$\theta$ = Angle of hit = $30^{\circ}$

x = Displacement in x direction = 400 ft

y = Displacement in y direction = 15 ft

$y_0$ = Height of hit = 2.5 ft

$a_y$ = g = Acceleration due to gravity = $32.2\ \text{ft/s}^2$

t = Time taken

Displacement in x direction

$x=u_xt\\\Rightarrow x=u\cos\theta t\\\Rightarrow t=\dfrac{x}{u\cos\theta}\\\Rightarrow t=\dfrac{400}{u\cos30^{\circ}}\\\Rightarrow t=\dfrac{400}{u\dfrac{\sqrt{3}}{2}}\\\Rightarrow t=\dfrac{800}{u\sqrt{3}}$

Displacement in y direction

$y=y_0+u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow y=y_0+u\sin\theta t+\dfrac{1}{2}a_yt^2\\\Rightarrow 15=2.5+u\sin30^{\circ}(\dfrac{800}{u\sqrt{3}})+\dfrac{1}{2}\times -32.2\times (\dfrac{800}{u\sqrt{3}})^2\\\Rightarrow \dfrac{400}{\sqrt{3}}-\dfrac{10304000}{3u^2}-12.5=0\\\Rightarrow 218.44=\dfrac{10304000}{3u^2}\\\Rightarrow u=\sqrt{\dfrac{10304000}{3\times218.44}}\\\Rightarrow u=125.4\ \text{m/s}$

The minimum initial velocity needed for the ball to clear the fence is $125.4\ \text{m/s}$

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3 years ago

Write the equation in standard form of the line that

aleksklad [387]

Answer:

x+2y=0

Step-by-step explanation:

y-y1=m(x-x1)

y-(-2)=-1/2(x-4)

y+2=-1/2x+4/2

y+2=-1/2x+2

y-(-1/2x)=2-2

y+1/2x=0

1/2x+y=0

2(1/2x+y)=2(0)

x+2y=0

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3 years ago

Suppose each square on the grid shown is 1 cm by 1 cm. Estimate the area of the figure.

finlep [7]

Answer:

10cm cm2

Step-by-step explanation:

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3 years ago

What are the domain and range of the function f(x) = 4? {x| x is a real number}; {y| y > 0} {x| x > 4}; {y| y > 0} {x|

Crazy boy [7]

Answer:

The domain is {x:x is a real number}

The range is {y:y is only 4}

Step-by-step explanation:

The domain of the function is the values of x

The range of the function is the values of y

f(x) = 4 is represented graphically by a horizontal line // to x-axis from y = 4

Every point on the this line has different values of x but same value of y

ex: (-1 , 4) , (5 , 4) , ( 0.1 , 4) , ................

So f(x) = 4 is called constant function

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The answer to this is a

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