4. Degree of f(x) * (g(x) = x^4 * 2x^4 = 2x^8 = degree 8
you're a life saver.
3. Note that if b = -1/2 we have
(x^4 - 3x^2 + 2) - (1/2)(2x^4 -6x^2 + 2x - 1) =
(x^4 - 3x^2 + 2) - x^4 + 3x^2 - x + 1/2 =
-x + 2 + 1/2 =
- x + (5/2)x = degree 1
Thanks!
1. 5x^3 + x^3 = 6x^3
2. a non-zero constant multiplier doesn't change thedegree of a polynomial
So the LARGEST possible degree of f(x) + a * g(x) = 4
68-22 = 70 - x
46 = 70 -x subtract 70 from both sides of the equation
-24 = -x
x = 24
I know a couple
Note that (3,2) is on the graph of g
So ( 2.3) is on the inverse....so....
g-1 (2) = 3
For the second...let h(x) = y
So....get x by itself
y =2x + 3
y - 3 =2x
(y -3) / 2 = x "swap" x and y
y = ( x - 3) / 2 = h-1(x)
Very nice solution, EP !!!
I had my second COVID shot yesterday....it obviously DIDN'T help my IQ much....LOL!!!!
Maybe I SHOULD breathe paint fumes !!
You are correct there ....and I can't solve them with my limited skills either !
x + 3y = - 5
3y = -x -5
y = -1/3 x - 5/3 slope = - 1/3 parallel slope is the same - 1/3
point slope form
(y-2) = - 1/3 ( x - -7)
y -2 = - 1/3 x - 7/3
y = - 1/3 x - 1/3 multiply by 3 to get rid of the fractions
3 y = -x -1 re-arrange
3y + x = -1 now multiply by -3
- 9y - 3x = 3 take it from here
Was thinking that.
Then went,
"Ah, must be some supremely difficult math equation I have no chance of solving."
Something is missing in your equations....there is no variable !
\(\text{GCF}(10p^2, 25pq, 15q^2) = \boxed{5}\)
To check, 10p^2/5 = 2p^2, 25pq/5 = 5pq, and 15q^2/5 = 3q^2, all of which have no factors in common.
Really, man?
sigh
Alright...
\(\sqrt92\)\(\approx\)9.59166
Yay!
I had my COVID vaccine this morning and have been breathing paint fumes the rest of the day.....that's what I'm gonna use for an excuse !!!!! Haha
thanx Varxaax !
Well, I can't know for sure, but I have a distinct feeling that it's 5.
I mean I'm pretty sure unless I have some numbers for p and q theres no way.
So uhh, either gimme some numbers, or live with the fact that it's most likely 5.
d / ( r + e ) = 6 d / ( r- e ) = 30 e = escalator speed r = run speed
30 r - 30 e = d
6r + 6e = d or 6e = d - 6r <====== multiply by -5 and sub into top equation
d = 30r-5d + 30 r
6d = 60 r
60 / 6 = d/r = 10 seconds Varaax is correct ......I fell prey to my warning about time in distance calcs !
Find x, y, and z such that x3+y3+z3=k, for each k from 1 to 100
Yikes. I feel bad for you, you gotta do some hard math, Mr. Guest.
I think you can do some rooting.
x+y+z=\(\sqrt[3]{k}\)
Other than that, I have 0 clue on how to keep going.
Will check....thanx V
Process of elimination.
The first one doesn't make sense, it's like cutting a giant hole in the back of a sweater.
Second one the long lanky things are hugging the shape-oid thing.
Third one you might get wrong but if you notice, it says six for the side panel ones.
So, it's D.
Note that she covers 1/18 mi in one minute
So
Rate * time = distance
(1/18) mi/mi * 42 min = 7/3 mi = total distance
(7/3) - (7/8) - (5/6) = 5/8 mi = .625 mi = distance we need
I don't think that is correct...
Shouldn't it be
2*(30/6)=10?
EDIT: oh no ep has been typing for quite a while i might get embarrassed.
EDIT 2: OH GOD NOW CPHILL IS HERE IMMA GET FLAMED
Just use the tangent inverse
arctan ( 350 / 1000) = the angle ≈ 19.2° = 19°
This is kinda like the boat and current speed questions
(6 + 30)/ 2 = speed on still escalator = 18 seconds
I'm literally so sad that I missed that.
