A) Isolate y in both inequalities
1) x + y ≥ 4 => y ≥ 4 - x
2) y < 2x - 3
B) Draw the lines for the following equalities:
1) y = 4 - x
2) y = 2x - 3
C) Shade the regions of solutions
1) The region that is over the line y = 4 - x
2) The region that is below the line y = 2x - 3
The solution is the intersection of both regions; this is the sector between both lines that is to the right of the intersection point, including the portion of the very line y = 4 - x and excluding the portion of the very line y = 2x - 3
The number of students in 5th grade is 60, since 24% of 250 is 60.
Since 90 is 38% if 250, 38 is going to be the percent of all students in sixth grade.
20% of 250 is 50, so the percent of all students in seventh grade is 50.
Finally, 45 is 18% of 250, so 18 percent of all students are in eight grade.
Hope this helped :)
17) f(x) = 16/(13-x).
In order to find domain, we need to set denominator expression equal to 0 and solve for x.
And that would be excluded value of domain.
13-x =0
Adding x on both sides, we get
13-x +x = x.
13=x.
Therefore, domain is All real numbers except 13.
18).f(x) = (x-4)(x+9)/(x^2-1).
In order to find the vertical asymptote, set denominator equal to 0 and solve for x.
x^2 -1 = 0
x^2 -1^2 = 0.
Factoring out
(x-1)(x+1) =0.
x-1=0 and x+1 =0.
x=1 and x=-1.
Therefore, Vertical asymptote would be
x=1 and x=-1
19) f(x) = (7x^2-3x-9)/(2x^2-4x+5)
We have degrees of numberator and denominator are same.
Therefore, Horizontal asymptote is the fraction of leading coefficents.
That is 7/2.
20) f(x)=(x^2+3x-2)/(x-2).
The degree of numerator is 2 and degree of denominator is 1.
2>1.
Degree of numerator > degree of denominator .
Therefore, there would no any Horizontal asymptote.
Answer: 68400
Explanation: The question is asking for area so you do base times height. In this case you multiply 360 times 190.
Answer:
70%
Step-by-step explanation:
