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Mumz [18]
3 years ago
6

What is a solution to a system of inequality's?

Mathematics
2 answers:
Damm [24]3 years ago
8 0
Make all equal or try to even it out depending on what the system is about
Reptile [31]3 years ago
8 0

Answer:

You can graph a linear inequality in two variables such as x or y, first get y solo on one side. The solution of the structure of inequalities is the intersection region of all the solutions in the system.

Step-by-step explanation:

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A) how many 3/5s are in 1<br> b) use the answer to part A to find the number of 3/5s in 2
BabaBlast [244]
a)\frac{1}{\frac{3}{5}}=1*\frac{5}{3}=\frac{5}{3}\\\\&#10;There\ are\ 1\frac{2}{3}\ of\ \frac{3}{5}s\ in\ 1.\\\\&#10;b)\frac{2}{\frac{3}{5}}=2*\frac{5}{3}=\frac{10}{3}\\\\There\ are\ 3\frac{1}{3}\ of\ \frac{3}{5}s\ in\ 2.
5 0
3 years ago
Read 2 more answers
Write an equation of the line that contains
Elodia [21]
Y-(-1)=3(x-3)->y+1=3x-9->y=3x-10 so y=3x-10 is your new line
7 0
3 years ago
Which is the correct answer?
Montano1993 [528]

Answer:

I'm not sure but I think it is letter A

Step-by-step explanation:

Because first off 0 is divisible by any integer number meaning it has to be divide by a whole number not fraction. Also, 0 + 2 = 2 then you divide that by 4 is 0.5.

6 0
3 years ago
A certain square is to be drawn on a coordinate plane. One of the vertices must be on the origin, and the square is to have an a
Scrat [10]

Answer:

The answer is (C) 8

Step-by-step explanation:

First, let's calculate the length of the side of the square.

A_{square}=a^2, where a is the length of the side. Now, let's try to build the square. First we need to find a point which distance from (0, 0) is 10. For this, we can use the distance formula in the plane:

d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} which for x_1=0 and y_1 = 0 transforms as  d=\sqrt{(x_2)^2 + (y_2)^2}. The first point we are looking for is connected to the origin and therefore, its components will form a right triangle in which, the Pythagoras theorem holds, see the first attached figure. Then, x_2, y_2 and 10 are a Pythagorean triple. From this, x_2= 6 or  x_2=8 while y_2= 6 or y_2=8. This leads us with the set of coordinates:

(\pm 6, \pm 8) and (\pm 8, \pm 6).  (A)

The next step is to find the coordinates of points that lie on lines which are perpendicular to the lines that joins the origin of the coordinate system with the set of points given in (A):

Let's do this for the point (6, 8).

The equation of the line that join the point (6, 8) with the origin (0, 0) has the equation y = mx +n, however, we only need to find its slope in order to find a perpendicular line to it. Thus,

m = \frac{y_2-y_1}{x_2-x_1} \\m =  \frac{8-0}{6-0} \\m = 8/6

Then, a perpendicular line has an slope m_{\bot} = -\frac{1}{m} = -\frac{6}{8} (perpendicularity condition of two lines). With the equation of the slope of the perpendicular line and the given point (6, 8), together with the equation of the distance we can form a system of equations to find the coordinates of two points that lie on this perpendicular line.

m_{\bot}=\frac{6}{8} = \frac{8-y}{6-x}\\ 6(6-x)+8(8-y)=0  (1)

d^2 = \sqrt{(y_o-y)^2+(x_o-x)^2} \\(10)^2=\sqrt{(8-y)^2+(6-x)^2}\\100 = \sqrt{(8-y)^2+(6-x)^2}   (2)

This system has solutions in the coordinates (-2, 14) and (14, 2). Until here, we have three vertices of the square. Let's now find the fourth one in the same way we found the third one using the point (14,2). A line perpendicular to the line that joins the point (6, 8) and (14, 2) has an slope m = 8/6 based on the perpendicularity condition. Thus, we can form the system:

\frac{8}{6} =\frac{2-y}{14-x} \\8(14-x) - 6(2-y) = 0  (1)

100 = \sqrt{(14-x)^2+(2-y)^2}  (2)

with solution the coordinates (8, -6) and (20, 10). If you draw a line joining the coordinates (0, 0), (6, 8), (14, 2) and (8, -6) you will get one of the squares that fulfill the conditions of the problem. By repeating this process with the coordinates in (A), the following squares are found:

  • (0, 0), (6, 8), (14, 2), (8, -6)
  • (0, 0), (8, 6), (14, -2), (6, -8)
  • (0, 0), (-6, 8), (-14, 2), (-8, -6)
  • (0, 0), (-8, 6), (-14, -2), (-6, -8)

Now, notice that the equation of distance between the two points separated a distance of 10 has the trivial solution (\pm10, 0) and  (0, \pm10). By combining this points we get the following squares:

  • (0, 0), (10, 0), (10, 10), (0, 10)
  • (0, 0), (0, 10), (-10, 10), (-10, 0)
  • (0, 0), (-10, 0), (-10, -10), (0, -10)
  • (0, 0), (0, -10), (-10, -10), (10, 0)

See the attached second attached figure. Therefore, 8 squares can be drawn  

8 0
3 years ago
HELP NEEEEED HELP !( POINTS PLEASE RIGHT ANSWER???????????????
Ivan

Answer:

Lydia

Step-by-step explanation:

5 0
2 years ago
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