Test - Final TEKS Check
for connexus, does anyone have all the answers ?

A sports magazine prints 12 issues per year, and a technology magazine prints 10 issues per year.
The total number of pages in all the issues of the sports magazine for one year is 32 more than the
1.
total number of pages in all the issues of the technology magazine for one year. Each issue of the
sports magazine has 18 fewer pages than each issue of the technology magazine. Which system of
equations can be used to find s, the number of pages in each issue of the sports magazine, and t,
the number of pages in each issue of the technology magazine?
S = t - 18
A.
12s = 10t + 32
t =s - 18
B.
10t = 12s + 32
S = t - 18
10s = 12t + 32
t =s - 18
D.
12t = 10s + 32
C.

the correct answer is b. negative exponential growth.

step-by-step explanation:

you can tell it's an exponential function because it decreases rapidly then levels out and never touches the x-axis.

it is negative because the y-axis decreases as the x-axis increases.

therefore it is a negative exponential growth graph.

hope i !

exactly you go it

step-by-step explanation:

13. jbhgfsfyfvhvbjko

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.