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I think it's 27.77 or 27.8

When you imput it into the calculator, you do this

18/ (24(1-0.7)^3)    I hope this helped.

I believe that every units in liters (especially milliliters) is the same amount in grams.

By using the change-of-base method......for example....suppose we have

log₃ 7     ........the answer to this is given by:     log(7) / log(3)  ≈ 1.77

So......as long as a, b > 0    and a not equal to 1, we have

log_a b =     log(b) / log(a)

Example: What is log of 15 to base 3:

First you take log(15) to base 10=1.176

Next you take log(3) to base 10  =.477

Then you just divide log(15) by log(3), thus:

1.176/.477=2.4654, which is log(15) to base 3, because,

3^2.4654=15-very nearly, because of rounding off.

What is the mass of 100mL?=100 Grams.

Just subtract the smaller amount from the bigger amount and that will give you what you want.

how do i calculate 5 % out of 580 000

5% means: 5/100=.05, then you just multiply by 580,000 like this:

.05 X 580,000=29,000-which is what you want.

40x1/4=40 X 1/4=40/4, divide both top and bottom by 4 and you get:

=10

(1+.003)(1+.013)(1+.097)^.33-1, As it is written, the answer is:

=0.0475592...........etc.

 v = 180(1 − 0.834^t)

Q: How long does it take to reach 65% of the terminal velocity? (Round your answer to two decimal places.)

65% of the termiinal velocity is: .65 X 180=117, then we have,

117=180(1 - .834^t), solving for t we get,

t=5.78 seconds - the time to reach 65% of terminal velocity.

Hallo guest

Hallo MM

50+123x13+345=50 + 1599 + 345=1994

Please write your question clearly, so that we can help you. Thanks.

Use this formula to solve your problem: D=1/2*g*t^2, where D=distance that the object has fallen, t=time in seconds, g=32 feet/sec^2-being the pull of gravity.

N = 39.5/

0.23 + 0.76^t foxes

Does the above mean: N=39.5/ (.23 + .76^t)? If yes, then we have:

95=39.5/(.23 + .76^t), and solving for t, we get:

t=6.13 Years for the population of foxes to reach 95.

Let X be the the intersection point between S and B

I assume you need surface area and volume

To get both you will need length SX

and to get surface area you will also need RS

\(tan30 = \frac{SX}{RX}\\\\ \frac{1}{\sqrt{3}}= \frac{SX}{12}\\\\ \frac{12}{\sqrt{3}}= SX\\\\ SX=4\sqrt{3}\;\; dm\\\\\)

\(cos30=\frac{RX}{RS}\\\\ \frac{\sqrt{3}}{2}=\frac{12}{RS}\\\\ \frac{2}{\sqrt{3}}=\frac{RS}{12}\\\\ \frac{24}{\sqrt{3}}=RS\\\\ RS=8\sqrt{3}\;\;dm\)

Can you take it from there?

If not just ask for more help :)

Oh you need to check my working too.  I think it is right but I don't put too much effort into proof reading because I think that is mainly your job.  If something doesn't make sense or you need more explanation, just let me know.  :)

Do you mean: 2Pi radians?. If so, then it just means: 2 X Pi=6.28318530.....radians=Cicumference of a circle in radians.

In a Mathematics test, Bala scored 5/6 of Calvin's score and Calvin scored 8 more marks than Dan. If the average score of the 3 boys is 88, what is Calvin's score?

Calvin's score=96

Bala's score  =80

Dan's score  =88

-30.774

The orbits of the two commets should coincide again in the year 2505 A.D.

(5^3)^4*5^3/5^2=(125)^4 X (125)/25=244,140,625 X 5

=1,220,703,125

single bar cost 35 pence special offer buy two bars get a nother one free a pack of 5 bar of chocholate costs £1.20 i want to buy five bars which is the cheapest way of buying them

Well, you can buy 5 in a pack for 1.20 pounds

or

You can buy 2 and get the 3rd one for free and then buy 2 more. = buy 4 =  4*0.35 = 1.40 pounds

BUT if you buy 4 you can get 2 for free so you might as well pay the 1.40 pounds and get 6 chopcolate bars.  If you are not hungry you can give one away or keep it for later.

So you can either pay 1.20 pounds for 5

OR

you can pay an extra 20p and get an extra one. (if they are all loose)

If you only want 5 it is cheaper just to buy the pack of 5 in the first place fo 1.20 pounds   :))

Josephine had a total of 540 Singapore and Malaysian stamps in the ratio 11:7. After she had given away an equal number of each type of stamps, the number of Singapore and Malaysia stamps left was in the ratio 7:1. How many Singapore stamps did she give away?

If you are up for it Maia I will work with you to solve this.  

Please other people do not jump in and give Maia the answer to this.  

She has told me that she want to learn and not just copy so this is a bit of a test. (You can answer her other questions if you want to although it would be good if you so not give her the whole answer all at once.)

---------------------------------------

Lets look at the first sentence.

Josephine had a total of 540 Singapore and Malaysian stamps in the ratio 11:7

This means that Josephine could seperate her Singapore and Malaysian stamps into two sperate piles.  She could then split her Singapore stamps into 11 separate piles and she could split her Malaysian stamps into 7 piles.  NOW each pile has exactly the same number of stamps.  Altogether there are 11+7=18 equal piles of stamps in front of her.

How many stamps are in each pile?  How did you get your answer?

Now.

How many of the stamps are Singapore ones and how many are Malaysian ones?  Show working.

Now it is time to interact with me Maia.  :))

Hallo MM