Answer:
the number to the right of the 7
Step-by-step explanation:
Answer:
7 ; 6
Step-by-step explanation:
Answer:
B.
R (2,4)
soln
center(h,k)=(6,1)
radius (r)=5units
so
eqn of circle is (x-6)^2+(y-1)^2=5^2
2 2
or, 25=x +y -12x-2y+36+1 now , only the given point (6,1)satisfies the eqn so this point lies on circle
E + w = 50.....w = 50 - e
25e + 20w = 1090
25e + 20(50 - e) = 1090
25e + 1000 - 20e = 1090
25e - 20e = 1090 - 1000
5e = 90
e = 90/5
e = 18 <=== the east got 18
w = 50 - e
w = 50 - 18
w = 32 <=== the west got 32
Answer:
Vertical A @ x=3 and x=1
Horizontal A nowhere since degree on top is higher than degree on bottom
Slant A @ y=x-1
Step-by-step explanation:
I'm going to look for vertical first:
I'm going to factor the bottom first: (x-3)(x-1)
So we have possible vertical asymptotes at x=3 and at x=1
To check I'm going to see if (x-3) is a factor of the top by plugging in 3 and seeing if I receive 0 (If I receive 0 then x=3 gives me a hole)
3^3-5(3)^2+4(3)-25=-31 so it isn't a factor of the top so you have a vertical asymptote at x=3
Let's check x=1
1^3-5(1)^2+4(1)-25=-25 so we have a vertical asymptote at x=1 also
There is no horizontal asymptote because degree of top is bigger than degree of bottom
There is a slant asympote because the degree of top is one more than degree of bottom (We can find this by doing long division)
x -1
--------------------------------------------------
x^2-4x+3 | x^3-5x^2+4x-25
- ( x^3-4x^2+3x)
--------------------------------
-x^2 +x -25
- (-x^2+4x-3)
---------------------
-3x-22
So the slant asymptote is to x-1
