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Anonymous
The process in which a solid solute leaves a solution and turns back into its solid phase (often noticeable in the bottom of the beaker) is called ______________.

The process is called precipitation.

-Frauke

Anonymous
what must be the same before two radicals can be added or subtracted? multiplied or divided?

For adding and subtracting, they need the same denominator (aka the number on the bottom).
For multiplication or division, they don’t need to be the same.

- Kendra

Anonymous
I did a maths exam and I got the hardest algebraic fraction I've ever gotten; 3x2 - 16x -35 over 9x2-25 There is no product and sum of 35 and 16, inc plus 4(or Atleast I couldn't find any) and there's no number I could take out between 9 and 25... It was also a non calculator paper, how do I solve the question(

Hi (:

Okay, the way you put the equation was a bit ambiguous but I interpret it as this:

$\frac{3x^2-16x-35}{9x^2-25}$

You are correct there’s no integer sum and product of 35 and 16 BUT you are forgetting about the 3 being the coefficient of x^2. so it will turn out as (3x + a)(x + b). Though 35 is still the product of a and b, the sum would be a + 3b. Looking at -35, the only factors available are 5 and 7 where one of them must be a minus.  if we do 5 - 3*7 we get -16 which is exactly what we’re trying to find. So we can change the above equation to:

$\frac{(3x+5)(x-7)}{9x^2-25}$

now let’s just simplify the denominator, as the coefficient of x is 0 we know that a and for this are equal with opposite signs, as the square root of 25 is 5 we can write the full equation as:

$\frac{(3x+5)(x-7)}{(3x+5)(3x-5)}$

cancelling some brackets we get:

$\frac{(x-7)}{(3x-5)}$

so x=7

- Sam

Anonymous
may I please have a step by step method to factor hard quadratic equations when the coefficient of x2 isn't 1..? I have a mock on wed😭

Hey Anon! We have a great step-by-step masterpost about quadratic expressions here, and it also covers factoring. There is also a great tutorial here which also covers factorising by solving first.

In short: Look for common factors/coefficents, and if this turns out to be too difficult, take the save road and use the quadratic formula.

-sorrel

Anonymous
Well holy. I had maths exam today and I didn't know how to do. This was one of the question. Solve the equation. (x-1)^2 = 7

x- 1 = sqrt(7)
x = 1 + sqrt(7)

:)

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Anonymous
Hello :) As i asked about the recent simultaneous linear. I've only learn two kinds which is Substitution and elimination only. how did you get 2t = 4 - 3w to t = 2 - 3w/2 ? isn't it suppose to be t = 4-3w/2 ? My exam is coming out next week! oh god. I learned all of these last year (I got an A got finals). I totally forgot how to do maths after the long term break from Nov to Dec. well, crap. :/

2t = 4 - 3w
so to get rid of the 2t and isolate it for just t we divide EVERYTHING by 2.
so the longer version is:

2t/2 = 4/2 - 3w/2
t = 2 - 3w/2

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Anonymous
Given that p-r/m = p/m . express p? with working pleaseeee

since both sides are over m, you cancel it out by multiplying both sides by m, leaving you with:
pm - r = p
pm - p = r
p(m- 1) = r
p = r/(m-1)

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Anonymous
4k^2 = 21 + 8k. Solve the equation

Hi (:

First let’s rearrange it into this form:

4k^2 - 8k - 21 = 0

Now if we remember factorising quadratics into brackets so they are in the form

(2k + a)(2k + b) = 0

we know the last number (in this case -21) is a multiplied with b

21 doesn’t have very many factors but by my guess is gonna be that a=±7 and b=3 where one of these numbers needs to be negative as it is -21

We also know that -8 is a sum of 2 lots of a added to 2 lots of as both the k values have coefficients of 2, as to factorise the 4k^2 down

so that’s ∓ 2x7 ± 2x3 if a = -7 then -14 + 6 = -8 so we know a is equal to -7. now we know (2k - 7)(2k + 3) = 0

So for f(k) to equal 0, k must equal 7/2 OR -3/2

(sorry, I got a little carried away explaining every single point but I hope it was helpful)

- Sam

Anonymous
I have a mock maths exam next week and I have lots to revise..I need to understand surds but I'm confused by it, and I don't want to spend all my time focusing on just surds and ignoring everything else! I understand the simple rules but struggle applying it to questions-could you perhaps explain them to me please? Gcse higher level x

Surd is a term that is used to talk about irrational numbers, especially regarding to square roots. A surd is the square root of something you can’t further simplify in the rational system (basically, it can’t be simplified down to a whole number). √2 is a surd, because it can’t be simplified in the rational system (you’d get 1.4142blablabla) and so is √3, because you’d get 1.7320blablabla.
On the other hand, √4 is NOT a surd, because it can become 2, which is still a rational number.
A surd is considered to be in its”simplest form” if the number under the root sign has no perfect square as a factor. (√a^2b can also be wrote as a√b)

The general rules you have to remember are: √ab=(√a*√b) and (√a*√a)=a, which can also be put as √a^2=a

Here  you can find a video on how to simplify surds, it’s divided into parts but they are all linked in the suggestions!

These are the major points that come to mind, if there is anything else just ask :)

Anonymous
I forgot how to do simultaneous linear equation. I needs find t and w. The equation is 3t-4w=23 and 2t+3w=4

There are many ways of solving these equations, but generally 3 known methods of solving these (to my knowledge) are: using substitution, elimination, and matrices.
The last one is a little bit similar to elimination but not quite useful with only 2 equations.

Substitution requires you to solve for one variable of one equation and then plug that value into the other equation. Ex:

3t-4w = 23
2t +3w = 4 —- We’ll use this one since it’s easier. Isolate for either t or w.

2t = 4 - 3w
t = 2 - 3w/2

Now the underlined part is what you’re going to put in for t in the other equation. What this does is it completely removed one variable from the above equation by substituting it’s value, and you’ll be left with only the w variable, You can continue from there to solve for w, and then use that solution for one of the variables and plug it back into either equation to solve for the remaining variable! :)

3(2 -3w/2) -4 = 23
6 - 9w/2 -4 =23/
9w/2 = 21
9w = 42
w = 42/9

2t + 3(42/9) = 4
2t + 13 = 4
2t = -9
t = -9/2

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