To solve this question, we use simple division:
So, there are
10 servings of 4/5 of a cup in 8 cups.
Answer:
Step-by-step explanation:
<u><em>The picture of the question in the attached figure</em></u>
we know that
The perimeter of a triangle is the sum of the length of their three sides
so
where
a,b,c are the length sides of the triangle
In this problem we have
substitute and solve for the missing length
Answer:
ask the goggle not 1 you help in thi aap bro your aand me same thing make sorry
Part A;
There are many system of inequalities that can be created such that only contain points C and F in the overlapping shaded regions.
Any system of inequalities which is satisfied by (2, 2) and (3, 4) but is not stisfied by <span>(-3, -4), (-4, 3), (1, -2) and (5, -4) can serve.
An example of such system of equation is
x > 0
y > 0
The system of equation above represent all the points in the first quadrant of the coordinate system.
The area above the x-axis and to the right of the y-axis is shaded.
Part 2:
It can be verified that points C and F are solutions to the system of inequalities above by substituting the coordinates of points C and F into the system of equations and see whether they are true.
Substituting C(2, 2) into the system we have:
2 > 0
2 > 0
as can be seen the two inequalities above are true, hence point C is a solution to the set of inequalities.
Part C:
Given that </span><span>Natalie
can only attend a school in her designated zone and that Natalie's zone is
defined by y < −2x + 2.
To identify the schools that
Natalie is allowed to attend, we substitute the coordinates of the points A to F into the inequality defining Natalie's zone.
For point A(-3, -4): -4 < -2(-3) + 2; -4 < 6 + 2; -4 < 8 which is true
For point B(-4, 3): 3 < -2(-4) + 2; 3 < 8 + 2; 3 < 10 which is true
For point C(2, 2): 2 < -2(2) + 2; 2 < -4 + 2; 2 < -2 which is false
For point D(1, -2): -2 < -2(1) + 2; -2 < -2 + 2; -2 < 0 which is true
For point E(5, -4): -4 < -2(5) + 2; -4 < -10 + 2; -4 < -8 which is false
For point F(3, 4): 4 < -2(3) + 2; 4 < -6 + 2; 4 < -4 which is false
Therefore, the schools that Natalie is allowed to attend are the schools at point A, B and D.
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