What is 1000 rounded to the nearest ten?

Chen has an MP3 player called the Jumble. The Jumble randomly selects a song for the user to listen to. Chen's Jumble has 4 clas

pochemuha

About 10 songs because if you add up all ten numbers of all the songs then you divide, then you divide again you will get 10 songs

6 0

3 years ago

Find the equation for the line that is parallel to the line with the equation y=−14x−1y=−14x−1 passing through the point (2,0)(2

bezimeni [28]

Use formula: y-y1/x-x1=m(gradient)
so, y=-14x-1
m=-14
y-0/x-2=14/1
y=-14x+28

7 0

3 years ago

Which equation describes the line parallel to y= -1/3x + 5?

bulgar [2K]

Answer:

its b i did that question on a test

Step-by-step explanation:

8 0

3 years ago

Please help :)

Degger [83]

Answer:

189 feet

Step-by-step explanation:

the distance from the drop zone the paratrooper will land = h × tan 8°

= 1350 × 0.14 = 189 feet

6 0

3 years ago

A steady wind blows a kite due west. The kite's height above ground from horizontal position x = 0 to x = 80 ft is given by y =

GaryK [48]

Answer:

The distance travelled by the kite is 122.8 ft ( approx )

Step-by-step explanation:

Here, the given function,

$y=150-\frac{1}{40}(x-50)^2$

Differentiating with respect to x,

$y'=-\frac{1}{20}(x-50)$

∵ arc length of a curve is,

$L=\int_{a}^{b} \sqrt{1+y'^2}dx$

Where, y shows the height of the curve for a ≤ x ≤ b,

Thus, the arc length of the given curve is,

$L=\int_{0}^{80} \sqrt{1+(-\frac{1}{20}(x-50)^2}dx$

Put $-\frac{1}{20}(x-50)=tan\theta$

$\implies -dx=-20 sec^2\theta d\theta$

$\implies L=-20\int_{0}^{80} \sqrt{1+tan^2\theta}sec^2\theta d\theta$

$=-20\int_{0}^{80} (sec \theta ) sec^2\theta d\theta$

By integration by parts,

$=|-\frac{20}{2}(sec \theta tan\theta +ln|sec\theta +tan\theta |) |^{x=80}_{x=0}$

If x = 80, $tan \theta = -\frac{1}{20}(30-50)=\frac{3}{2}$

$sec \theta = \frac{\sqrt{13}}{2}$

$\implies \theta = \frac{1}{20}(0-50)=\frac{5}{2}$

$sec \theta = \frac{\sqrt{29}}{2}$

Thus, the length of the curve is,

$=-10(\frac{\sqrt{13}}{2}(-\frac{3}{2}) +ln|\frac{13}{2}-\frac{3}{2}|) + 10(\frac{5\sqrt{29}}{4} + ln |\frac{29}{2} + \frac{5}{2} |)$

$\approx 122.8\text{ feet}$

8 0

3 years ago