7. You and a friend are paid $38.25 for doing yard

Response:

b

Detailed explanation:

Answer:

50 Educators

Step-by-step explanation:

To tackle this question, the initial step is to calculate the amount of teachers prior to the addition of new staff. For this, I devised Model 1. In this model, teachers are positioned at the top of the ratio and students at the bottom. The variable X represents the number of teachers we are determining. Utilizing this model, I computed 2,100 multiplied by 1 (2,100) and then divided by 14 to conclude there were 150 teachers. Next, I formed a similar model with the updated student-teacher ratio (Model 2). This time, I multiplied 2,100 by 2 (which is 4,200) and divided by 21 to ascertain there are 200 teachers. Having established both the initial and the increased counts of educators, subtracting the original from the new gives you the tally of new teachers, which results in an increase of 50 teachers.

Step-by-step explanation:

A = {0,1,2,3}

a): R = {(0,0),(2,2),(3,3)}

R displays antisymmetry, as whenever (a,b)∈R, it follows that a=b.

R lacks reflexivity since (1,1) ∉ R even though 1 ∈ A.

R is transitive; therefore, if (a,b)∈R and (b, c) ∈ R, then a=b=c and (a,c)=(a,a)∈R.

R fails to be a partial ordering due to its lack of reflexivity.

b): R = {(0,0),(1,1),(2,0),(2,2),(2,3),(3,3)}

R is antisymmetric because if (a,b)∈R and (b, a) ∈ R, then a must equal b (e.g., (2,0) ∈ R and (0,2) ∉ R; likewise, (2,3) ∈ R and (3,2) ∉ R).

R is reflexive since each (a,a) resides in R for all elements a ∈ A.

R is transitive; if (a,b)∈R and (b,c)∈R, it implies (a,c) exists in R or identical to (a,b) in R.

R qualifies as a partial ordering due to its reflexivity, antisymmetry, and transitivity.

c): R =  {(0,0),(1,1),(1,2),(2,2),(3,1),(3,3)}

R is reflexive as (a,a)∈R is true for every a ∈ A.

R is antisymmetric; if (a,b)∈R holds and if also (b,a)∈R, then a invariably equals b (e.g., (1,2)∈R while (2,1) ∉ R; similarly for (3,1) and (1,3)).  

R fails transitivity because (3,1) ∈ R and (1,2) ∈ R, but (3,2) ∉ R.

R is not a partial ordering due to transitivity not being satisfied.

d): R =  {(0,0),(1,1),(1,2),(1,3),(2,0),(2,2),(2,3), (3,0),(3,3)}

R exhibits reflexivity since (a,a)∈R for each element a ∈ A.

R displays antisymmetry, as if (a,b)∈R and (b,a)∈R then a must equal b (e.g., (1,2)∈R and (2,1)∉R; similarly validated for others).

R is not transitive because (1,2)∈R and (2,0)∈R, but (1,0)∉R.

R is not a partial ordering due to transitivity issues.

e):  R = { ( 0, 0 ), ( 0, 1 ), ( 0, 2 ), ( 0, 3 ), ( 1, 0 ), ( 1, 1 ), ( 1, 2 ), ( 1, 3 ), ( 2, 0 ), ( 2, 2 ), ( 3, 3 ) }

R proves to be reflexive, given that (a,a)∈R for all a∈A.

R is not antisymmetric since both (1,0)∈R and (0,1)∈R hold while 0 is distinct from 1.

R lacks transitivity, as (2,0)∈R and (0,3)∈R, while (2,3)∉R.

R cannot be classified as a partial ordering as it fails in both antisymmetry and transitivity.

Take note of the image below

the riverbank requires no fencing due to the river's presence
so the pen's perimeter can be calculated as 2w + l, or w + w + l
thus   

derive P(w), set it to zero, locate any critical points, and perform a first-derivative test for minimum values.

To achieve the desired output, first use the machine with the function y = x^2 - 6, followed by the machine that computes y = sqrt(x-5). This way, when you input 6, the output from the first machine is calculated as x = 6, yielding y = 6^2 - 6, resulting in 30 as the input for the second machine. The second machine then processes this to provide the final output of sqrt(30 - 5), which equals sqrt(25) = 5. Alternatively, to obtain a negative final output, you should first utilize the machine with the function y = sqrt(x-5). Assuming you start with the value x = 9, the first machine computes this to sqrt(9-5), which is sqrt(4) = 2. Then, the second machine converts y to 2^2 - 6, leading to a result of 4 - 6 = -2.