Answer:
The answer is 16.
Step-by-step explanation:
Answer:
2
g
n
+
3
+
3
g
Step-by-step explanation:
Answer:
48 uniforms
Step-by-step explanation:
1836-612=1224.
1224/25.50=48.
Checked answer:
25.50 x 48= 1224
If the circle has the same center as the diagonals of a square and the radius of the circle is smaller than 1/2 the diagonal of the square but larger than 1/2 the length of the side of a square, then there are 8 points of intersection -- 2 at each corner of the square.
If the radius of the circle is smaller than 1/2 the side length of the square and the center is as described above, there are no points of intersection.
If the circle is located outside the square it can have 1 tangent point or 2 intersection points depending on the location conditions of the circle in relation to the square.
Answer:
We show that f(x) n+8/6n = 6 x n = 0
which flips the n+8/1 = 0+8/0-6= x = 3 this is the range.
For the HA we would work left to right.
x goes to positive or negative infinity and is determined by the highest degree terms of the polynomials in the numerator and the denominator. This particular function has polynomials of degree 0 in both the numerator and the denominator
If say n+8 was n+2 then we would use the 2/-2+3 and get 1 and show the hole as the source;
hole : -2+1 as non equal sign. but not in the case of n+8/6n
-2+1 represents 1/3 symmetry.
We see for n+8/6n with interpreted back into the zero format minus
-0+8/-0-6 we see there is symmetry and can work on the left side of graph and flip over. Where 0 = n+8 and 1=nx6
Step-by-step explanation:
There would be no way of doing the others unless the exponents had been squared ^2
If they were squared then the domain will be (-infinity -3) parenthesis
union of( -3 -2) union of +2 to negative infinity.
There is not a vertical asymptote as the numerator divides into dominator at point 8 as a decimal.
The holes are then closed.
