5. Note that since p(x) has degree 11 and q(x) = 7
Then p(x) + g(x) will have a degree of 11....in other words.....the degree of q(x) is irrelevant in the addition
4. Degree of f(x) * (g(x) = x^4 * 2x^4 = 2x^8 = degree 8
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
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
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 !!
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
Just use the tangent inverse
arctan ( 350 / 1000) = the angle ≈ 19.2° = 19°
At 12K per hour....it takes him 5 min per each kilometer (do you see this ??? )
(Think about it.....there are 12 equal parts in one hour ...so...he runs 2 km in 10 min )
So....he has to cover 10 - 2 = 8 kilometers in 50 remaining minutes = 5/6 hr
Rate = Distance / Time
Rate = 8 /( 5/6) = 8 * (6/5) = 48/5 km /hr = 9.6 km/ hr for the rest of the race
Orange Dr = sqrt (85^2 - 77^2) = sqrt (1296) = 36
77 + 36 = 113
113 - 85 = 28 yds farther
