Things you have to remember Flashcards
What should you ensure if you’re dealing with a question involving time, distance and speed?
All time is converted to hours or any units the question mentioned.
E.g., 36 minutes for km/h needs to be converted by 36/60, thus 0.6
What should you do if you get these types of questions?
x/2 = x/6
The denominator multiplies the other side of the equation numerator, and the same is true for the other.
How about
x/2 = x/6 - x/3?
Find the LCM
Make sure they all have LCD then remove it.
then do your usual algebra
How do you do these questions?
36/3x-6 + 1/x-2 = 13/3
- You find the LCD
Notice how 3x-6 can be 3(x-2)? All other denominators also have it
- Times 3(x-2) to all fractions,s and from there you should be good to go
36/3x+6 * 3(x-2) will become 36 because like terms. Put it in a fraction and you would cross them out, otherwise it will just equal to *1.
1/x-2 * 3(x-2) would be 3 because you cancel x-2 and you would be left with 1*3
13/3 * 3(x-2) would be 13(x-2) because 3 and 3 cancels.
When moving a denominator to mutliply the other side, what should you make sure?
eg x/2 + 2 = 2
x + 4 = 4
you must times the number near it as well.
what is the quadratic formula?
x = (-b ± √(b^2 - 4ac)) / 2a
In a number line, is |-17| negative?
No, -|17| is otherwise is a positive 17.
What do you do when you get these types of questions?
sqrt(25y^9/x^6)
The rule of a square root (or radical) to an exponent is that it halves it.
In this case y^9 becomes y^9/2 which is y^4 sqrt(y) since there is 1 remainder.
When the exponents are odd, split it up by 2 and make the number of remainders in a radical.
so this case
sqrt(25) = 5
sqrt(y^9) = y^4 sqrt(y)
sqrt(x^6) = x^3
thus
5y^4sqrt(y)/ x^3
What to do you in these types of questions?
6sqrt(112) + 6(175)
You break down each radical into an “squarable number” and make sure they all have the same radical expression.
112 / 2 = 56, a non-squarable number, so wrong.
Squarable numbers are 4,9,16,25,36,49,64
112 / 4 = 28. Since 4 is a square number, can you do the same with the next expression, 175? No, so you can’t use 4.
Keep testing in the calculator
Find the common radical. Keep doing so because that is the most simplified; if you can’t, then use an uncommon radical expression
- 112/16 = 7
- 175/25 = 7 as well.
There, you found the golden number
sqrt(16) sqrt(7) = 4 sqrt(7), thus 6(4sqrt(7)) thus 24sqrt(7)
Do the same with 2, and it should look like this
24sqrt(7) + 30sqrt(7)
= 54sqrt(7).
If it is uncommon, separate them like
5sqrt(7) + 6sqrt(3)
Solve the quadratic equation by factoring
49x - 42x^2 = 0
No c? No problem. Cringe.
Okay but fr
Order it around to
42x^2 - 49x = 0
Now fun the GCF (greatest common factor).
This case is 7x. Then factor it, with 7x outside the paraentheses.
7x()
What makes 42x^2? 6x
What makes 49x? 7
So
7x(6x-7) = 0
Split it up
7x = 0.
x = 0
6x-7 = 0
x = 7/6
thus
0, 7/6 is the answer.
solve by using the square root property
(x-7)^2 = 16
move ^2 to 16
x-7 = 4, -4
x = 11 , 3.
z^2 - 14z +13 = 0
Solve by completing square
- move 13
z^2 - 14z = -13
- For some reason, half -14 and square it
1/2 * -14 = -7, -7^2 = 49
- Add 49
z^2 - 14z + 49 = 36 (added 49 to -13)
then you factor 49 and all that
z(z-7)-7(z-7)
(z-7)(z-7) turn into (z-7)^2
(z-7)^2 = 36
- Move ^2 to 36, algebraically becoming square root thus 6
- z-7 = 6, -6
get the answer of both
thus 13,1.
Once again… solve the quadratic equation below by completing the square.
z^2 + 3z - 54 = 0
- Move 54
z^2 + 3z = 54
- Half 3 then sqaure it
z^2 + 3z +9/4 = 54 + 9/4, into 225/4
- see that z^2? move that squared into a square root to 9/4 and stuff 3z, you get 3/2
- arrange
(z+3/2)^2 = 225/4
- move exponent across into root
z+3/2 = 7.5
find z, -z whatever.
What is the formula used to solve quadratic equations?
a^2 + bx + c = 0
Use that FIRST in any question.
IF THAT DOESNT WORK, do the root one.
How do you solve
4x^2 + 11x = 3??
See that 4? divide it to 11 and 3.
Now do the rest, you should seen it in the other flashcards.
In differentiation calculus, how can you find the derivatives of x^3?
d/dx [x^3] = 3x^2. Because you have to move the exponent to the front and -1 the exponent.
In differentiation calculus, how can you find the derivatives of 3?
d/dx [3] = 0. Is always 0.
In differentiation calculus, how can you implement the chain rule and when?
Use this when you have a function within a function like f(g(x)).
You would then change that to
dy/dx = deratives of f’ (g(x)) TIMES deratives of g’(x)
so dy/dx = f” (g(x)) * g”(x)
In differentiation calculus, how can you implement the product rule and when?
You use the product rule when you’re multiplying 2 functions. Think that it doesn’t have multiplication or a function within a function.
Formula :
y = u(x)⋅v(x)
becomes
dy/dx = deratives of u (x) * v(x) + u(x) * deratives of v (x)
soo dy/dx = u”(x) * v(x) + u(x) v”(x).
You basically find the derivative for u THEN you leave the rest, add same but derivatives of another variable which is v.
In differentiation calculus, how can you implement the quotient rule and when?
ANY TIME YOU SEE A FRACTION OR DIVIDE.
so if y = u(x)/v(x)
you do
dy/dx = v(x) * the derative of u(x) - u(x) * the deratives of v(x) OVER / / / (v(x))^2.
Denominator SQAURED.
so
dy/dx = v(x) * u”(x) - u(x) * v”(x) / (v(x))^2
When do you use log and ln?
log_5 125 = 3
meaning 5^3 = 125.
Solving log would look like
log_5 y = log_5 2 + log_5 x
in log, + means times and log_5, treat them as like terms
so
log_5 y = log_5 2x
5^log(y) = 5^log(2x)
Both side 5^log so cross out
Result
y = 2x.
When do you use exponential?
if you see log_5 125 = 3
Just get that number in log base, 5 and put 3 from the other side as an exponent of 5.
so 5^3 = 125.
What is the formula for series?
a_n = ar^n-1
a = number
n = place of number (1,2,3 etc, the question would ask you to find places)
r means rate of the series, to calculate that you get the n/n-1
for eg a_2/a_1
so if you had a pattern of 3,6,9
r = 6/3 = 2. (this is arithmetic but the question will give you a geometric with a constant rate).
What the hell is even differentiation?
The measuring of the speed of things changing in the moment on the curve graph.
