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3 years ago

How many positive integers less than 1000 (a) are divisible by 7? justify your answer?

1 answer:

mote1985 [20]3 years ago

6 0

142.

You can get this by seeing what number you have to multiply 7 by to get as close to 1000 as you can. 7*142 = 994. Thus each one of the multiplications up to that point will give you a positive integer than can be divided by 7. <span />

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Help me with this question pleaseeee

Lerok [7]

Answer:

a=69

Step-by-step explanation:

121-52 is equal to 69

3 0

3 years ago

Read 2 more answers

What is 224 divided by 60 in fraction form?

REY [17]

373/100 is 224 divided by 60 in a fraction form

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3 years ago

The probability that Scott will win his next tennis match is 3/5

goblinko [34]

Answer:

the probability that he will not win is 2/5 cause it's just the chance that he doesn't win

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3 years ago

if a motercycle is moving at a constant speed down the highway of 40 km/hr, how long would it the motorcycle to travel 10 km

ahrayia [7]

Answer:

15 minutes

Step-by-step explanation:

First, the motorcycle goes at a speed of 40 km/hr.

The question asks for how long it would take to travel 10 km.

Well, there are 60 minutes in an hour, since we will be translating into minutes.

Also, 10 km is 1/4 of 40 km, so it would make sense that the time length would be 1/4 of an hour as well.

1/4 of 60 minutes is 15 minutes. So it takes 15 minutes for the motorcycle to travel 10 km.

Now, if all this wordy stuff is too much to comprehend, you can also solve using proportional relationships.

$\frac{40km}{60min}=\frac{10km}{xmin}$

Now cross multiply:

$40km*xmin=10km*60min\\40x=600$

Divide both sides by 40:

$\frac{40x}{40}=\frac{600}{40}\\x=15$

Again, this shows that it wouls take 15 minutes for the motorcycle to travel 10 km.

6 0

3 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

3 years ago