Mathematics, 30.01.2020 17:00 skylarleannjone2751

Which graph correctly represents 1/3y-1/2x> 2 ?

Answers: 2

Show answers

Answers

Put it in the form of y =mx +b, or in this instance, y> mx +b
move the (1/2) x to the right by adding it to both sides of the inequality
(1/3)y>(1/2)x +2
Multiply by 3 on both sides to get y by itself.
y>(3/2) x +6
This is a graph with y intercept of (0,6) and a moderate upward and to the right slope. Because it is > , the line on the graph will NOT be part of the solution.

The easiest way to find the side of the graph that the inequality satisfies is to use (0,0) and see if it works or doesn't work. In the original equation, 0-0>2 does NOT work, so the area where the inequality works is to the up and left of the graph, which should be a dotted line to show that the inequality is greater than only.
The point (6,-2) should work.
Test it. 6*(1/3)-(-2)*(1/2)>0 ; 2-(-1)=3, and 3>2 It does work.
The point (6,2) should not work
Test it. 6 *(1/3)-2(1/2)=2-1 ; 1 is NOT >2, so it does not work.
If the graph goes through the origin, then pick a point near the graph with a small x or y.

Answer from: Quest

12 rio aoice of q kqpd so r lagr

Answer from: Quest

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

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

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

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.

Other questions on the subject: Mathematics

Mathematics, 21.06.2019 22:30, rivera6681

Solve: 25 points find the fifth term of an increasing geometric progression if the first term is equal to 7−3 √5 and each term (starting with the second) is equal to the difference of the term following it and the term preceding it.

Answers: 1

continue

Mathematics, 22.06.2019 00:20, sheena33

Aline is perpendicular to another line m2= (-1/m1), where m1 is the slope of the original line and m2 is the slope of the perpendicular line. which of the following equations represents the line that passes through the point (-2,1) and is perpendicular to the line below?

Answers: 2

continue

Mathematics, 22.06.2019 04:00, sammyraegarrett

The number of students who ride a school bus is 110% of the number of students who walk. how many students ride the school bus?

Answers: 1

continue

Mathematics, 22.06.2019 04:10, s945924

Choose the domain & range of the relation shown in the graph: domain: -1, 0, 1, 2, 3 range: -3, -1, 0, 3 domain: -3, -1, 0, 3 range: -3, -1, 0, 3 domain: -3, -1, 0, 3 range: -1, 0, 1, 2, 3 domain: 3, 1, 0, 3 range: -1, 0, 1, 2, 3