Please assist me with the domain and range of graphs

 see below

Step-by-step explanation:

Domain represents the x-values from the smallest (furthest left) to the biggest (furthest right).

Range represents the y-values from the lowest (furthest down) to the highest (furthest up).

Interval notation: If a value is included (closed dot), use a bracket [  ]

                      If a value is NOT included (open dot), use a parenthesis (  )

                     Note that ± ∞ is never included.

1. Domain: smallest x-value is -4 (included). biggest x-value is 3 (included)

  Range: lowest y-value is 2 (included). highest y-value is 5 (included)

                 D: x = [-4, 3]        R: y = [2, 5]

2. Domain: smallest x-value is -∞ (←). biggest x-value is 2 (included)

  Range: lowest y-value is 2 (included). highest y-value is ∞ (↑)

                 D: x = (-∞, 2]        R: y = [2, ∞)

3. Domain: smallest x-value is -∞ (←). biggest x-value is ∞ (→)

  Range: lowest y-value is -∞ (↓). highest y-value is ∞ (↑)

                 D: x = (-∞, ∞)        R: y = (-∞, ∞)

4. Domain: smallest x-value is -∞ (←). biggest x-value is ∞ (→)

  Range: lowest y-value is -∞ (↓). highest y-value is 4 (included)

                 D: x = (-∞, ∞)        R: y = (-∞, 4]

5. Domain: smallest x-value is -∞ (←). biggest x-value is ∞ (→)

  Range: lowest y-value is 3 (included). highest y-value is ∞ (↑)

                 D: x = (-∞, ∞)        R: y = [3, ∞)

6. Domain: smallest x-value is -2 (included). biggest x-value is ∞ (→)

  Range: lowest y-value is -3 (included). highest y-value is ∞ (↑)

                 D: x = [-2, ∞)        R: y = [-3, ∞)

answer: ratio of volume of cone (with 3 times the height and equal diameter) to volume of cylinder is 2 to 1. the value of π is not used.

step-by-step explanation:

radius is diameter/2 = 4.

area of circle is π r^2.

volume of cylinder is height × area of base = hπr^2 = 3(4^2) π = 48π cubic inches.

not knowing the formula for volume of cone, i calculate it using calculus (which is for calculating things).

volume of cone is integral 0 to h area(y) dy. area(y) is π r(y)^2. r(y) runs r to zero as y runs zero to h, r(0) = r, r(h) = 0.

r(y) = r - r/h y

area(y) = π (r - r/h y)^2 = π (r^2 - 2r^2y/h + r^2y^2/h^2) = π r^2(1-2y/h+y^2/h^2)

antiderivative area(y) = π r^2 (y-y^2/h+y^3/(3h^2))

integral 0 to h = π r^2(h-h^2/h+h^3/(3h^2) - (0)) = π r^2(h - h + h/3) = π r^2(h/3) = π/3 r^2 h

volume of this cone is π/3(4^2)18 cubic inches.

= 6 16 π = 96π cubic inches.

cone has twice the volume, 96/48.

cone height is 6 times the cylinder.

ratio of volume formula: π r^2 cancels.

cone over cylinder ratio is 6h/3 to h, 2 to 1.

delaney would like to make a 5lb nut mixture that is 60% peanuts and 40% almonds.

she has several pounds of peanuts and several pounds of a mixture that is 20% peanuts and 80% almonds.

let 'p' represent the number of pounds of peanuts needed to make the new mixture, and  

let 'm' represent the number of pounds of the 80% almond-20% peanut mixture.

:

p + m = 5

therefore

m = (5-p)

write an equation based on the percent of peanuts

1p + .2m = .6(5)

replace m with (5-p)

p + .2(5-p) = .6(5)

p + 1 - .2p = 3

p - .2p = 3 - 1

.8p = 2

p = 2/.8

p = 2.5 lbs of peanuts required

then

5 - 2.5 = 2.5 lbs of the peanut/almond mixture required

:

a) what is the system that models this situation?

p + m = 5

1p + .2m = .6(5)