right, silly me..

so that would give us 3y=6
y=2?

ja das correct

1)When x=1 the expression should be 5
2)Not defined for x=8

15x-20/x-8

I don't understand this whatsoever..

another example

when x=0 the expression should be 2
not defined for x=3

answer should be: x-6/x-3=-6/-3=2... but I don't know how

don't know if the exact translation is that from swedish --> english but let me know if anything is unclear..

1 more:

x-6/2x^2+10x+12

we start by putting =0 in the denomineter.

2x^2+10x+12=0

now how to go from here? I don't know how I'm supposed to calc it.

Sorry didn’t see it was a rational. I’ll let someone else handle this

This post was edited by brigadier on Jan 17 2019 09:07am

Sure no problem, don't be sorry. Mistakes can happen since I'm not good at explaining stuffs anyway
I'll wait for a proper answer to my question(s)

1)
This is asking you to substitute 1 for x and simplify

You have

(15(1)-20)/(1-8)

(15-20)/(-7)

(-5)/(-7)

5/7



2) probably asking you to realize that dividing my zero is undefined depending on the textbook you could set the denominator not equal to zero and solve for any undefined values OR plug in 8 and show that you would divide by zero

First method
Set x-8=\=0
X=\=8

Second method

(15(8)-20)/(8-8)

100/0 <—— we generally don’t write division by zero but we see plugging in 8 would give you this undefined value.

What is the question asking exactly?

You have a few methods for solving
2x^2+10x+12=0

I would suggest factoring and getting the practice.

I would start by factoring out a 2 from each term because I realize they are all even and divisible by 2

2(1x^2 +5x+6)=0

You can now factor the quadratic (or use the quadratic formula)

We want factors of ac=(1)(6)=6 that add to b=5

We list factor pairs until we find one that satisfies
1,6 (does not add to 5)
2,3 (2*3=6 and 2+3=5)

We can then rewrite the quadratic with 2x+3x instead of 5x

we have

2(X^2 +2x + 3x + 6)=0

Group the first and last two terms

2((x^2 +2x )+ (3x + 6))=0

Factor out x from the first paraenthesis and 3 from the second to get

2( x (x+2) + 3(x+2) )=0

We can rewrite as
2 (x+3)(x+2)=0

We can use the zero product property to set each oaraenthesis equal to zero

X+3=0. X+2=0

X=-3. X=-2. These would be your two undefined values or your vertical asymptotes depending on what you are actually trying to find.

Thank you @ above.

Go a physics problem here.

A car that drives at 49km/h is approaching a distance with the speed limit 30 km/h. The soft braking lasts for 2.5 seconds.

1) How big is the deceleration?

2) How far before the sign does the driver atleast have started slowing down?

So we would use the formula: v=v0+at and get that

v0=49km/h but it has to be converted to m/s so that's (49*1000)/3600 = 13.6m/s

And the speed limit sign (30km/h) also can be converted to m/s (30*1000)/3600 = 8.3m/s

Right and our "t" was stated in the question (2.5 seconds)

So if we slap the numbers into the aforementioned formula we get:

8.3=13.6+a*2.5

13.6 moved over to left side gives 8.3-13.6=a*2.5

Now I'm stuck, don't know how to go from here.. anyone that knows? I need to get a=-2.1m/s^2 somehow

if a*2.5=8.3-13.6 then chances are a=(8.3-13.6)/2.5 ~=-2.1 (m.s^(-2))

also whats up with these quadratic equations? seems like youve been at it for yrs. its literally alwayyyys the same thing... do you ever learn?

are you mentally challenged or something? are you trolling? not meant to be a flame im just wondering how somebody can own a porsche (https://forums.d2jsp.org/img.php?x=390567) (i.e. an adult) and still struggle with this

either way why are you studying mathematics if you are so bad at it? most jobs dont require mathematics, its a pretty useless way to spend ones time, many will disagree but im just saying

good luck