360÷40=36 tens÷4 tens

Mathematics

2 answers:

Thepotemich [5.8K]3 years ago

7 0

360 / 40 = 36 right, so if it is 36 tens which = 360 and 4 tens = 40. Your answer is 9 :D

Effectus [21]3 years ago

5 0

Your Answer Is " 9 "

You might be interested in

Find the diameter and of each circle. A) circumference = 118mm B) circumference = 1m

kotegsom [21]

Okay so the formula for circumference is

C=(pi)(D)
or
C=(pi)(2)(r)

But since they gave us the diameter we are gonna have to use the first formula.

Now I don’t know if they want u to use 3.14 for pi or the actual number, so i’m gonna assume we are using the actual formula.

So the formula to be able to find the diameter of circle A is going to be:
118/pi.
So put that into your calculator and you get 37.5605mm or 37.56mm.
It depends on what place your instructor wants you to round to.

The formula for circle B is going to be:
1/pi.
So let us put that into our calculator and you get 0.318309mm etc. Or just 0.3183.

I believe that I am correct but if i’m not please correct me.

4 0

3 years ago

WHY IS NOBODY HELP ME I GOT 2 OTHER QUESTIONS POSTED AND NOBODY ANSWERED ME

andriy [413]

Not everybody knows the answers and sees your questions

3 0

2 years ago

Read 2 more answers

Find the radius of convergence of the power series. (if you need to use or –, enter infinity or –infinity, respectively. ) [infi

Alona [7]

For given power series $\sum_{n=0}^{\infty} \frac{(-1)^nx^n}{2^n}$ the radius of convergence is 2.

For given question,

We have been given a power series $\sum_{n=0}^{\infty} \frac{(-1)^nx^n}{2^n}$

We need to find the radius of convergence of the power series.

We use ratio test to find the radius of convergence of the power series.

Let $a_n=\frac{(-1)^nx^n}{2^n}$

$\Rightarrow a_{n+1}=\frac{(-1)^{n+1}x^{n+1}}{2^{n+1}}$

Consider,

$\lim_{n \to \infty}|\frac{a_{n+1}}{a_n} |\\\\= \lim_{n \to \infty} |\frac{\frac{(-1)^{n+1}x^{n+1}}{2^{n+1}}}{ \frac{(-1)^nx^n}{2^n} } |\\\\=\lim_{n \to \infty} |\frac{(-1)^{n+1}x^{n+1}}{2^{n+1}}\times \frac{2^n}{(-1)^nx^n} |\\\\=\lim_{n \to \infty} |\frac{(-1)x}{2} |\\\\=\lim_{n \to \infty}|\frac{-x}{2} |\\\\=\frac{x}{2}$