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2 years ago

The graph of f(x) = x2 + x - 6 is shown below.

Mathematics

1 answer:

guapka [62]2 years ago

3 0

Answer:

the solutions of a function are the points where for some value of x the function becomes zero

thus the solns for this graph would be

that's option 1.

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Solve each equation. 0.2x2 + 1.2 = 3.4

Marat540 [252]

$0.2x^2+1.2=3.4\ \ \ |-1.2\\\\0.2x^2=2.2\ \ \ |:0.2\\\\x^2=11\to x=\pm\sqrt{11}\\\\\text{Answer:}\ x=-\sqrt{11}\ \vee\ x=\sqrt{11}$

6 0

3 years ago

Read 2 more answers

P is a polynomial of degree 4 with three zeros: -1 and 2 are zeros of multiplicity one, and 3 is a zero with multiplicity two. I

tatyana61 [14]

Answer:

$P(x) = (x+1)(x-2)(x-3)^2 + 108$

Step-by-step explanation:

let me know if you want an explanation

3 0

1 year ago

Find the work done by F= (x^2+y)i + (y^2+x)j +(ze^z)k over the following path from (4,0,0) to (4,0,4)

babunello [35]

$\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k$

We want to find $f(x,y,z)$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=x^2+y$

$\dfrac{\partial f}{\partial y}=y^2+x$

$\dfrac{\partial f}{\partial z}=ze^z$

Integrating both sides of the latter equation with respect to $z$ tells us

$f(x,y,z)=e^z(z-1)+g(x,y)$

and differentiating with respect to $x$ gives

$x^2+y=\dfrac{\partial g}{\partial x}$

Integrating both sides with respect to $x$ gives

$g(x,y)=\dfrac{x^3}3+xy+h(y)$

Then

$f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)$

and differentiating both sides with respect to $y$ gives

$y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C$

So the scalar potential function is

$\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}$

By the fundamental theorem of calculus, the work done by $\vec F$ along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it $L$ ) in part (a) is

$\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}$

and $\vec F$ does the same amount of work over both of the other paths.

In part (b), I don't know what is meant by "df/dt for F"...

In part (c), you're asked to find the work over the 2 parts (call them $L_1$ and $L_2$ ) of the given path. Using the fundamental theorem makes this trivial: