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3 years ago

Plz help thx appreciate

Mathematics

1 answer:

lozanna [386]3 years ago

4 0

P= 12 + 4x^2 - 2 + x^2 + 3x^2
Combine like terms
12-2= 10
4x^2 + 3x^2 + x^2= 8x^2
P= 8x^2 + 10
Sorry if this isn't right but this is what I think you would do

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Gwen says that the sum of -1 3/4 and 2 1/2 is the same as the difference between 2 1/2 and 1 3/4. Is Gwen correct? Explain why o

solong [7]

Answer: No

Step-by-step explanation: no because one is negative 1 3/4 and the other one isint in the negatives

4 0

2 years ago

Point N(7, 4) is translated 5 units up.

lutik1710 [3]

It is given in the question that

Point N(7, 4) is translated 5 units up.

And we have to find the coordinates of its image after this transformation.

Since the point translated up by 5 units, so the x coordinate remains same and we have to add 5 to y coordinate.

So the new coordinate after the transformation is

$N'(7+5,4)
\\
N'(12,4)$

And that's the required coordinate after the given transformation .

5 0

2 years ago

What are the factors of the equation x 2 - 5x + 6?

xxTIMURxx [149]

Answer:

(x-3)(x-2)

Step-by-step explanation:

x^2 -5x+6

What two numbers multiply to 6 and add to -5

-3 * -2 = 6

-3 -2 = -5

(x-3)(x-2)

7 0

2 years ago

Please help me ASAP

stellarik [79]

Answer:

<em>The Graph is shown below</em>

Step-by-step explanation:

<u>The Graph of a Function</u>

Given the function:

$\displaystyle y=g(x)=-\frac{3}{2}(x-2)^2$

It's required to plot the graph of g(x). Let's give x some values:

x={-2,0,2,4,6}

And calculate the values of y:

$\displaystyle y=g(-2)=-\frac{3}{2}(-2-2)^2=-\frac{3}{2}(-4)^2==-\frac{3}{2}*16=-24$

Point (-2,-24)

$\displaystyle y=g(0)=-\frac{3}{2}(0-2)^2=-\frac{3}{2}(-2)^2=-\frac{3}{2}*4=-6$

Point (0,-6)

$\displaystyle y=g(2)=-\frac{3}{2}(2-2)^2=-\frac{3}{2}(0)^2=0$

Point (2,0)

$\displaystyle y=g(4)=-\frac{3}{2}(4-2)^2=-\frac{3}{2}(2)^2=-\frac{3}{2}*4=-6$

Point (4,-6)

$\displaystyle y=g(6)=-\frac{3}{2}(6-2)^2=-\frac{3}{2}(4)^2=-\frac{3}{2}*16=-24$

Point (6,-24)

The graph is shown in the image below

8 0

2 years 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:

$\displaystyle\int_{L_1}\vec F\cdot\mathrm d\vec r=f(0,0,0)-f(4,0,0)=-\frac{64}3$

$\displaystyle\int_{L_2}\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(0,0,0)=\frac{67}3+3e^4$

8 0

2 years ago