The graph of f(x) = 2x is shown on the grid. On a coordinate plane, an exponential function approaches y = 0 in quadrant 2 and increases into quadrant 1. It goes through points (0, 1), (1, 2), (2, 4).

The graph of g(x) = (One-half)x is the graph of f(x) = 2x reflected over the y-axis. Which graph represents g(x)?

On a coordinate plane, an exponential function approaches x = 0 in quadrant 4 and increases into quadrant 1.
On a coordinate plane, an exponential function decreases from quadrant 2 into quadrant 1 and approaches y = 0.
On a coordinate plane, an exponential function approaches y = 0 in quadrant 3 and curves down into quadrant 4.
On a coordinate plane, an exponential function approaches x = 0 in quadrant 1 and curves down and to the right into quadrant 4.

On a coordinate plane, an exponential function approaches y = 0 in quadrant 2 and increases into quadrant 1. It goes through points (0, 1),

sorry i dont really know

Step-by-step explanation:

the answer is:

the answer is:

to solve the problem, we need to remember the quotient of power property, it's defined by the following relation:

if we have a quotienf of powers that have the same base, we need to keep the same base and subtract the exponent of the denominator power to the exponent of the numerator power.

so, we are given the expression:

then, calculating we have:

hence, the answer is:

have a nice day!

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.