The equation for the motion of an object with constant acceleration is d=vt+0.5at? where d is distance traveled in ft, v is starting velocity in ft/s, a is acceleration in ft/s2, and t is time in seconds. A racecar begins with a velocity of 200 ft/s and accelerates at 105 ft/s2. Write an expression for the distance the car travels after t seconds. What is the GCF of the expression? ​

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.

system of linear equation

step-by-step explanation:

as given in the data

a= 20 , b= -12

let us suppose that it is a straight line equation

now we know the formula for the system of linear equation

b= ma + c )

here m is slope of a straight line which is equal to formula

m=   (ii)

and here c is a constant

now putting the value of a and b in equation (ii)

m=

m=

putting the value of a,b and m in equation (i) gives us

-12 = + c

-12=-12 + c

adding 12 to both sides

-12 +12 = -12 +12 +c

0 =c

or c=0

as the value of the constant is zero which gives us that it is a linear equation and this order pair satisfies the system of linear equations.