All categories

3 years ago

What is the square root off 60000000 These teachers trippin

Mathematics

2 answers:

garri49 [273]3 years ago

7 0

7745.96669241 is the square root...hope it helps :)

skelet666 [1.2K]3 years ago

5 0

that would be 7745.96 from my best guess

You might be interested in

The first term in the expansion of the binomial (3x + y)4

Lelechka [254]

I don't know if i'm right or what because I haven't learned binomial in like almost 4 years. (3x+y)?

5 0

3 years ago

Read 2 more answers

Find the perimeter

Inga [223]

Answer:

Step-by-step explanation:

Perimeter= 4S

Perimeter= 4(5)

Perimeter= 20

Perimeter 2= 2(L+W)

Perimeter 2= 2(11+7)

Perimeter 2= 22+14

Perimeter 2= 36

Total Perimeter= P1+P2

Total Perimeter= 20+36

Total Perimeter= 56

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%2826%20%5Cdiv%20100%29%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5Ctimes%2010" id

taurus [48]

Answer:

$\boxed{\bf \: \cfrac{13}{5}}$

Or in Decimal:

$\boxed{\bf \: 2.6}$

Step-by-step explanation:

Given expression :-

$\sf \: ( 26 \div 100) \times 10$

Solution :-

$\sf = (26 \div 100 )\times 10$

This arithmetic expression may be rewritten as ;

$\sf = \cfrac{26}{100} \times 10$

Step 1 : Cancel the zero of 10 and one zero of 100 :-

$\sf = \cfrac{26}{10 \cancel0} \times 1 \cancel0$

Results to;

$\sf = \: \cfrac{26}{10} \times 1$

$\sf = \: \cfrac{26}{10}$

Step 2: Cancel 26 and 10 by 2 :-

$\sf = \cfrac{ \cancel{26}}{ \cancel{10}}$

Results to;

$\sf = \cfrac{ \cancel{26} {}^{13} }{ \cancel{10} {}^{5} }$

$\sf = \cfrac{13}{5}$

It can also be in Decimal.

That is;

$\sf = 2.6$

Hence, the answer of the expression would be 13/5 or 2.6 .

$\rule{225pt}{2pt}$

I hope this helps!

Let me know if you have any questions.

I am joyous to help!

3 0

3 years ago

Read 2 more answers

Is 0.9727 a positive correlation, negative correlation, or no correlation?

lapo4ka [179]

I think it’s positive correlation

4 0

3 years ago

1) Use power series to find the series solution to the differential equation y'+2y = 0 PLEASE SHOW ALL YOUR WORK, OR RISK LOSING

iogann1982 [59]

$y=\displaystyle\sum_{n=0}^\infty a_nx^n$

then

$y'=\displaystyle\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty(n+1)a_{n+1}x^n$

The ODE in terms of these series is

$\displaystyle\sum_{n=0}^\infty(n+1)a_{n+1}x^n+2\sum_{n=0}^\infty a_nx^n=0$

$\displaystyle\sum_{n=0}^\infty\bigg(a_{n+1}+2a_n\bigg)x^n=0$

$\implies\begin{cases}a_0=y(0)\\(n+1)a_{n+1}=-2a_n&\text{for }n\ge0\end{cases}$

We can solve the recurrence exactly by substitution:

$a_{n+1}=-\dfrac2{n+1}a_n=\dfrac{2^2}{(n+1)n}a_{n-1}=-\dfrac{2^3}{(n+1)n(n-1)}a_{n-2}=\cdots=\dfrac{(-2)^{n+1}}{(n+1)!}a_0$

$\implies a_n=\dfrac{(-2)^n}{n!}a_0$

So the ODE has solution

$y(x)=\displaystyle a_0\sum_{n=0}^\infty\frac{(-2x)^n}{n!}$

which you may recognize as the power series of the exponential function. Then

$\boxed{y(x)=a_0e^{-2x}}$

7 0

3 years ago