@jim_thompson5910 If (x-2)^2=25 and x<0, what is the value of x? -23 -7 -5 -3 -2
take the square root of both sides don't forget the \(\pm\)
I squared both sides and solved for x but it wasn't there. I just hate the SAT math lol
You need to isolate x. So your first step is to undo that squaring exponent.
Ohhh right so it would be -3.
square root, not square
you don't square both sides you need to undo squaring (so square root)
yes
checking your answer is always a good idea Check: (x-2)^2=25 (-3-2)^2=25 ... replace every x with -3 (-5)^2 = 25 25 = 25 ... checks out
I know I meant square root it. See I just used +5 then added 2 to get 7 and just chose -7 because idk instead of using -5+2=-3
I dont work well when I'm being timed. Okay next one
Let's check x = -7 (x-2)^2=25 (-7-2)^2=25 (-9)^2 = 25 81 = 25 ... this is a false equation, so x = -7 is not the answer
even though you are in a timed setting, checking helps
but I can see how you got it since x = 7 works (x-2)^2=25 (7-2)^2=25 (5)^2=25 25 = 25
If m=t^3 for any positive integer t, and if w=m^2+m, what is w in terms of t? t^2+t t^3 t^3+t t^5+t^3 t^6+t^3
w=m^2+m is in terms of m we want the right side to be just t thankfully m is in terms of t since m = t^3
so wherever you see an 'm' you replace it with t^3 and simplify
Oh duh. Wow.
that's ok
t^6+t^3 I always misread and freak about about questions like that even though they are easy
just take your time and follow the set up gameplan
Its just they dont give us a lot of time lol
it's better to execute properly than go too fast
that's true, which is why you practice
if the problem seems too tough, then skip it and move onto other ones get those out of the way and you'll build up momentum to come back and finish the ones you skipped
just make sure you don't go too fast or take too long on one problem
Okay next one: (this one has xtriangle so delta x? So I'll use xt because I cant put a triangle) For all positive integers x, let xt be defined to be (x-1)(x+1). Which of the following is equal to 6t-5t I just put t there because there is no x just a triangle and the answer choices only have triangles and no x's.. 2t+1t 3t+2t 4t+3t 5t+4t 6t+5t
hmm odd notation so you have \(\Large x\Delta\) or \(\large \Delta x\) ?
Its x with a colored triangle after it. But that's only when it says let let xtriangle be defined etc etc. But in the 6-5 and all the answer choices there are just colored in triangles after each number
I'll post a picture
thanks
Its confusing to me too lol
the SAT is always confusing lol
I know so dumb
they're the masters at making things more complicated than they have to be
Exactly. -_-
thanks
No thank you lol
yep, I remember this cryptic bs...lol anyways, let's make the triangles 't's like you have it
since xt = (x-1)(x+1) this means 6t = (6-1)*(6+1) ... replaced every x with 6 (xt ---> 6t) 6t = 5*7 6t = 35
and 5t = (5-1)*(5+1) 5t = 4*6 5t = 24
So... 6t - 5t = 35 - 24 = 11
pic just loaded i remember doing this question from the collegeboard book
The final result of 6t - 5t is 11 so we have to find the answer choice that also gives you a result of 11
Ohhh okay so 6 and 5?
6t+5t?
you mean choice E?
let's find out 6t = 35 (work shown above) 5t = 24 (work shown above) which means 6t + 5t = 35+24 = 59 that's not equal to 11 like we want, so choice E is eliminated
Oh.. I was thinking (6-1)(5+6) oh I added them ugh oops
always stick to the definition xt = (x-1)*(x+1)
This problem is really confusing me I'm sorry... so how do I find the answer?
Let's check choice A
well we're looking for which answer gives us 11 ill try choice b
Oh wait I know how to find the answer now one second
xt = (x-1)(x+1) 2t = (2-1)(2+1) 2t = 1*3 2t = 3 --------------- xt = (x-1)(x+1) 1t = (1-1)(1+1) 1t = 0*2 1t = 0 --------------- so, 2t + 1t = 3 + 0 = 3 that means choice A is out
the good thing about multiple choice is that you can check each possible choice with the answer you want in this case, we want 11 to be the result so far, we haven't found it
It's b
good
you'll find that 3t = 8 and 2t = 3 so 3t + 2t = 8+3 = 11
it's a good idea to check the others, but I have a feeling they are wrong
Thanks sorry I'm kind of in a hurry it's getting late, but I understand that one now. If x^2/y is an integer, but x/y is not an integer, which of the following could be the values of x and y? x=1 y=1 x=3 y=2 x=4 y=2 x=6 y=4 x=9 y=3
I did check the others when I tried the others to get the answer lol
again we have a list of choices, so why not check them all
this is really just where you plug in the answers and see which answers fits what is asked
checking A x^2/y = (1^2)/1 = 1/1 = 1 ---> integer x/y = 1/1 = 1 ---> integer so A is out
A is out because we want x/y to NOT be an integer
So that means E and C are out too
good
nice job on eliminating answers
But B and D are both not integers. And I did all this while taking the test and got stuck there.
checking B x=3 y=2 x^2/y = (3^2)/2 = 9/2 = 4.5 ---> not an integer so B is out because x^2/y has to be an integer
Ohhh right I see
checking D x=6 y=4 x^2/y = (6^2)/4 = 36/4 = 9 x/y = 6/4 = 3/2 = 1.5 so that means D is the answer
x^2/y is an integer, while x/y is not
|dw:1402110942228:dw| not an integer anyways if you really get stuck just plug in each answer choice
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