3 years ago

HELP, it is important!

Mathematics

1 answer:

suter [353]3 years ago

5 0

Barrow Alaska. Its the lowest tempreture.

You might be interested in

11/15 - 1/4 = ????<br> no links please!

valkas [14]

Answer:0.483

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

If a chord 10 inches long is 5 inches from the center of a circle, find the radius of the circle.

Sedaia [141]

check the picture below.

6 0

3 years ago

Read 2 more answers

The park plans to hire 20 percent more than the minimum number of ticket sellers needed in order to account for sickness, vacati

Ira Lisetskai [31]

The park plans to hire 20 percent more than the minimum number of ticket sellers needed in order to account for sickness, vacation, and lunch breaks. How many ticket sellers should the park hire? Explain.

8 0

2 years ago

The line 6x – 3y = 24 represents Shari's distance from home, y in blocks, on her run across town over time, x, in minutes. What

Charra [1.4K]

Answer:

The x-intercept of the line is 4 minutes.

That 4 minutes gives the time when Shari will reach her home.

Step-by-step explanation:

The line 6x - 3y = 24......... (1), represents Shari's distance from home, y in blocks, on her run across town over time, x, in minutes.

So, at y = 0, from the above equation (1), we get 6x - 0 = 24, ⇒ x = 4.

Therefore, the x-intercept of the line (1) is 4 minutes.

That 4 minutes gives the time when Shari will reach her home and the distance between her and her home is zero blocks. (Answer)

4 0

3 years ago

Read 2 more answers

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