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3 years ago

How do you find two numbers whose quotient is about 4.1

1 answer:

3 0

So you're looking for 2 numbers you can divide to get ABOUT 4.1. This means you're going to estimate. First you can estimate that 4.1 is about 4 and then think what two numbers divide to give you 4? 8÷2 = 4 ... so now we can change it slightly to make the quotient a little higher and say
8.1÷2 = 4.05 That will round to 4.1 so you could say two numbers who's quotient is ABOUT 4.1 would be...

8.1 and 2

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Look at the bottom of the pictures to see what number it is

kondaur [170]

Step-by-step explanation:

hence the value of y=-9/4and k=-16.

3 0

2 years ago

Find NQ if NP = 13 cm and line l is a segment bisector.

Zielflug [23.3K]

Answer:

NQ = 26 cm

Step-by-step explanation:

A line bisector divides a line into two equal segments.

Given that line l is the bisector of segment NQ, it then means it divides NQ into two equal segments, namely, segment NP and segment PQ.

Thus, NP = ½ of segment NQ

Therefore, if NP = 13 cm,

13 = ½*NQ

multiply both sides by 2

2*13 = NQ

26 = NQ

NQ = 26 cm

7 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: