Use the work to complete the sentences describing the solution set to the compound inequality 2x + 9 < 17 and 14 < –7 + 3x. The graph of x < 4 is
_.
(Options are:
A.) A closed circle at 4 shaded to the right
B.) A closed circle at 4 shaded to the left
C.) An open circle at 4 shaded to the right
D.) An open circle at 4 shaded to the left)

The graph of 7 < x is
_.
(Options are:
A.) A closed circle at 7 shaded to the right
B.) A closed circle at 7 shaded to the left
C.) An open circle at 7 shaded to the right
D.) An open circle at 7 shaded to the left)

Because this is an “and” compound inequality, solutions satisfy
_.
(Options are:
A.) Both inequalities
B.) Either inequalities
C.) Neither inequalities)

The solution set to the compound inequality is
_.
(Options are:
A.) 4 < x < 7
B.) x < 4
C.) x < 7
D.) No solution)

phosphate, calcium ion, chloride ion, neon atom

step-by-step explanation:

step-by-step explanation:

. your answer will contain x - and y -terms. (1 point) x ( ¿¿ 2 − 7 ) 2 +( y 2 − 9 ) 2 √ ¿ 12. what is the distance from the directrix (the crocodile river) to the point ( x , y )? write this equation. your answer will contain a y -term.

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.