Carly is constructing the inscribed circle for △MNP . She has already used her compass and straightedge to complete the part of the construction shown in the figure.

What could be her next step?

Use the straightedge to draw the line through points J and K.

Use the straightedge to draw the line through points J and P.

Place the point of the compass on point M and draw an arc that passes through points N and P.

Place the point of the compass on point J and draw an arc in the interior of the triangle.
A triangle with vertices labeled as M N P. The top vertex is labeled as M. An arc is drawn through vertex N that intersects side M N and N P at point J and K, respectively.

he can buy 4lbs. of live bait and 1lbs of natural bait

step-by-step explanation:

6^2 or 36

step-by-step explanation:

6000n

6×1000n

6×10^3 n

so we need 6n need to be a perfect cube

so n has to be 6^2 so that 6n is 6^3

what i'm saying is if you let n be 6^2 you have 6^3×10^3 or 60^3 which is a perfect cube

mr. india

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.