All categories

3 years ago

What's 0.00045 in scientific notation

Mathematics

2 answers:

tino4ka555 [31]3 years ago

8 0

0.00045 = 4.5 × 10^-4

Vitek1552 [10]3 years ago

8 0

00045 becomes 4.5 x 10-4<span>, because we need to divide 4.5 by 10,000 to get it to become 0.00045</span>

You might be interested in

Chinese checkers is a game in which players move their pieces across a board using single moves or jumps like the traditional ga

DENIUS [597]

A. it is a hexagon (there are 6 sides) B. using the equation (n-2)180 (where n is the number of sides we get (6-2)180=4*180=720. divide 720 by 6 to get 120 for each angle of the hexagon. (it has no irregularities in it's shape so that number will be correct for them all.) C. to get that just divide 39 by 6 to get 6.5 in per side. Hope that helps:)

6 0

3 years ago

Please help me. please, I need help. Help me please.

Ierofanga [76]

A. The Square Root of 11 is 3.31662479(etc.)

8 0

3 years ago

Read 2 more answers

A collection of quarters and nickels contains at least 42 coins and is worth at most $8.00. If the collection contains 25 quarte

Elenna [48]

Answer:

Step-by-step explanation:

9 0

3 years ago

Read 2 more answers

the first ice cream was made april 3rf 8192 if there is 3 flavors 2 toppings and 3 fruits how many different sundae combinations

miv72 [106K]

Answer:

The answer to this problem is 18 combinations

Step-by-step explanation:

If you have 3 flavors, 2 toppings and 3 fruits the number of combos you can make is 18 because:

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Ctiny%5Cint_%7Be%7D%5E%7B%7Be%7D%5E%7B2%7D%7D%20%5Cleft%28%20%5Csqrt%7Bx%20%2B%20%5Cfrac%

dlinn [17]

We have the identity

$\left(\sqrt x + \dfrac1{2^n}\right)^2 = x + \dfrac1{2^{n-1}} \sqrt x + \dfrac1{2^{2n}}$

Take the square root of both sides and rearrange terms on the right to get

$\sqrt x + \dfrac1{2^n} = \sqrt{x + \dfrac1{2^{n-1}} \left(\sqrt{x} + \dfrac1{2^{n+1}}\right)}$

Decrementing n gives

$\sqrt x + \dfrac1{2^{n-1}} = \sqrt{x + \dfrac1{2^{n-2}} \left(\sqrt{x} + \dfrac1{2^{n}}\right)}$

and substituting the previous expression into this, we have

$\sqrt x + \dfrac1{2^{n-1}} = \sqrt{x + \dfrac1{2^{n-2}} \sqrt{x + \dfrac1{2^{n-1}} \left(\sqrt x + \dfrac1{2^{n+1}}\right) } }$

Continuing in this fashion, after k steps we would have

$\sqrt x + \dfrac1{2^{n-k}} = \sqrt{x + \dfrac1{2^{n-(k+1)}} \sqrt{x + \dfrac1{2^{n-k}} \sqrt{x + \dfrac1{2^{n-(k-1)}} \sqrt{\cdots \dfrac1{2^{n-1}} \left(\sqrt x + \dfrac1{2^{n+1}}\right)}}}}$

After a total of n - 2 steps, we arrive at

$\sqrt x + \dfrac14 = \sqrt{x + \dfrac12 \sqrt{x + \dfrac1{2^2} \sqrt{x + \dfrac1{2^3} \sqrt{\cdots \dfrac1{2^{n-1}} \left(\sqrt x + \dfrac1{2^{n+1}}\right)}}}}$

Then as n goes to infinity, the first nested radical converges to √x + 1/4. Similar reasoning can be used to show the other nested radical converges to √x - 1/4. Then the integral reduces to

$\displaystyle \int_e^{e^2} \left(\sqrt x - \frac14\right) + \left(\sqrt x + \frac14\right) \, dx = 2 \int_e^{e^2} \sqrt x \, dx = \boxed{\frac43 \left(e^3 - e^{\frac32}\right)}$

5 0

2 years ago