Please helppp with this

What is the period of the function y=3/2con(3/5x)+5

jeyben [28]

Answer:

$y = \frac{3}{2} cot( \frac{3x}{5} ) + 5 \\ \\ \pi \div \frac{3}{5} \\ \ \\ = \frac{5\pi}{3} is \: period$

6 0

2 years ago

I need help with this!?!

AlekseyPX

Some of the important "given" information is outside of the photo.
We need to know that the two triangles are similar.
And we need to know that the WHAT ? of angle M is 9/40.

7 0

2 years ago

What dose -9.5-(-8) - 6.5 equal

jarptica [38.1K]

Answer:

-24

Step-by-step explanation:

-9.5 - 8 would be -17.5. subtract 6 from that and you get -24, the answer.

8 0

3 years ago

Select the favorable outcomes for rolling doubles.

QveST [7]

By doubles as outcome means the first and second outcome must be same.

The first option has outcome as 1. (1-1) (2-2) (3-3) (4-4) (5-5) (6-6).

In this we have first and second outcome as same, so we can say it satisfies our requirement and option first is the correct choice.

8 0

2 years ago

Read 2 more answers

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

2 years ago

Please helppp with this​

Please helppp with this