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Hey Oscar, it looks like you've already answered the question! Usually we would work backwards to find out the nth term from a sequence of numbers.

In this case it looks like we would have had a sequence like:
-4, -2, 0, 2, 4 etc.
where this sequence starts from the 1st term

Detecting the difference between each number in the sequence and seeing that it 'goes up' by 2 each time, usually means the n is multiplied by 2, i.e. 2n

We can only assume the '-6' part of the equation if we were told that the 1st term is -4. If we are not told which nth the original sequence is, you may be being asked a different question, like to determine which term a number would be (what is the value of n) based on the number in the sequence and the given equation.

Hope this helps!

Maths

The nth term in the sequence is 2n - 6

After a bit of research I can see that in most cases this question means to present your answer as TWO to the power of something. (when expressed 'as a power of two') It can be tricky when we also hear 'TO the power of 2' or 'power 2'. 

In this case, 128 can be broken down into 2 multiplied by itself 7 times, AKA:
2 TO THE POWER OF SEVEN
2^7

add back in the Sq root.
√(2^7)

as i mentioned before, this can be written as (2^7)^0.5

Using the multiplication of exponents in brackets rule
(x^m)^n = x^mn,

We can multiply the powers together (the one inside the bracket and the one outside) so:
(2^7)^0.5
2^(7x0.5)

So in this case your answer would be 2^3.5

Write 
√
128
 as a power of 
2
.

Hi Oliwia, not sure if we are missing some information to complete this question, but I can share a couple of things that may help:

Any square root can be expressed as a power of 1/2, in this case:
√ 128 = 128^1/2 - though I think you want your answer expressed to the power of 2, not to the power of half.

If you were looking to have an answer that was to the power of 2, you may need to first solve √128, which is 11.31. If you want to express 11.31 as a power of 2, you would need to find the square root of this.
√11.31 = 3.36

3.36^2 is likely to be your answer based on the wording of your question.

Also note that we found out the square root of a square root, this doesn't mean we found the cubed root, and because we did this in stages, it would not work to reverse the calculations to get our original number.

Hope this helps!

Hi Dacoda!

I think we need a little more information to work out the true mass, though I can share with you that "mass = volume x density". Let's do the maths with Water, as with water 1 millilitre = 1 gram mass

So in this case:
volume = 1000ml
density = 1g/ml

mass = 1000 x 1
mass = 1000g

Hope that helps, be sure to find out the density when approaching a question like this. Usually expressed as 'g/ml' which tells you how many 'grams per millilitre'.

1 litre of a liquid P has a mass of P grams

Hi Dallas!
Anytime you are told that a value is fraction, and you are asked to find the full amount. One common method to solve this is to first find '1/?', where the numerator is 1.

In this case we want to find 1/3. So if know that 'two thirds of the value is 25' then we know that 'one third of the value is 25 - halved'. Because we can halve 2 to get one.

So 25/2 is equal to a third, to get the full value we need to multiply by 3 (to make the numerator match the denominator).

-> 25/2 x 3 = 75/2
(when multiplying a fraction by a whole number, you only multiply the numerator)

Your answer is 75/2 expressed as a fraction or 37.5!





If 2/3 of a number is 25 what is the number

First I would separate out the equations so it's easier to break down for yourself:
a+b-c = 5
b+c-a = 6
a+c-b = 7

There are a few ways to solve this, today lets use 'elimination'. This is where we will add two equations together and try to 'cancel out' certain letters or values.

let's look for a two equations which 'cancel out' something if we added them together. In the first equation there is a '-c' so lets add it to an equation that has '+c' (take the second equation for example)

So line up the two equations and rearrange them so that the values are neatly aligned:
 a+b-c = 5
-a+b+c = 6    <-this was originally "b+c-a = 6" 

If you look closely, it looks like both the 'a's and the 'c's will cancel out, now treat this like a simple addition, and leave the equals sign in!

  +a +b -c = 5
  -a +b +c = 6    +
---------------
0 + 2b + 0 = 11

so now we know that 2b = 11, we know that "b = 11/2" (or 5.5)

Let's go back and see if we can use this method to find another letter, we can apply it to the second and third equation:
b+c-a = 6
a+c-b = 7

the 'b's and 'a's should cancel out
   -a +b +c = 6
   +a -b +c = 7   +
---------------
0 + 0 + 2c = 13

now that we know 2c = 13, we know that c = 13/2 (or 6.5)

using 'b = 5.5' and 'c = 6.5' we can substitute these values into any of the equations to find 'a'.
 b + c - a = 6
 5.5 + 6.5 - a = 6
 12 - a = 6
 12 = 6 + a
 6 = a

So:
a = 6
b = 5.5
c = 6.5



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