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3 years ago

Mohjjbn bn hjbjn hjbnm

1 answer:

7 0

Yup yup, I understand your question: jejsusuehsbhs lakksjjwhe eiksjsjnesh is the answer you’re looking for. I’m happy to help

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Give your answer in terms of <img src="https://tex.z-dn.net/?f=%5Cpi%20%5C%5C" id="TexFormula1" title="\pi \\" alt="\pi \\" alig

Marina CMI [18]

The length of the band in terms of π is given by the expression:

L = 40mm + π*10mm

<h3>How to get the length of the band?</h3>

Remember that for a circle of diameter D, the circumference is:

C = π*D.

Now, if you look at the image, you can see that the length of the band will be equal to 4 times the diameter of the pencils (one time for each side). Plus 4 times one-fourth of the circumference of each pencil (for the four corners).

because the diameter is 10mm, the length of the band will be:

L = 4*10mm + 4*(π*10mm/4)

L = 40mm + π*10mm

This is the length of the band in terms of π.

If you want to learn more about circles:

brainly.com/question/14283575

#SPJ1

7 0

2 years ago

Can you tell me what 4 2/5 is as a fraction with dollar and cents sign?

amid [387]

$4.40 cents would be ur answer...

5 0

4 years ago

Read 2 more answers

Help please i need the answer.

kow [346]

i think you just have to do 3 + 14+2+2+3and then solve i guess

7 0

2 years ago

a 36 foot is attached from the top of a pole to a bracket that is 9 feet from the base of the pole how tall is the pole?

olya-2409 [2.1K]

The answer is 4feet

3 0

3 years ago

Consider a particle that moves through the force field F(x, y) = (y − x)i + xyj from the point (0, 0) to the point (0, 1) along

sleet_krkn [62]

The work done by $\vec F$ is

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r$

where $C$ is the given curve and $\vec r(t)$ is the given parameterization of $C$ . We have

$\mathrm d\vec r=\dfrac{\mathrm d\vec r}{\mathrm dt}\mathrm dt=k(1-2t)\,\vec\imath+\vec\jmath$

Then the work done by $\vec F$ is

$\displaystyle\int_0^1((t-kt(1-t))\,\vec\imath+kt^2(1-t)\,\vec\jmath)\cdot(k(1-2t)\,\vec\imath+\vec\jmath)\,\mathrm dt$

$=\displaystyle\int_0^1((k-k^2)t-(k-3k^2)t^2-(k+2k^2)t^3)\,\mathrm dt=-\frac k{12}$

In order for the work to be 1, we need to have $\boxed{k=-12}$ .

3 0

3 years ago