The range of all real numbers x such that 2x − 5 < 7 and 4x + 10 > 6 is . the range of all real numbers x such that 2x − 5 < 7 or 4x + 10 > 6 is .

Solution for Part 1:

2x − 5 < 7 and 4x + 10 > 6 is called a compound inequality. To find the range we need to solve each of the inequality and range would be the values of x which will make both the inequality TRUE.

Solving inequality 2x - 5 < 7:

2x - 5 < 7 {adding 5 on both sides}

2x - 5 + 5 < 7 + 5 {Combining like terms in either side of inequality}

2x < 12 {Dividing 2 on both sides}

2x/2 < 12/2 {Simplifying fraction in either side of inequality}

x < 6

Solving inequality 4x + 10 > 6:

4x + 10 > 6 {Subtracting 10 on both sides of inequality}

4x + 10 - 10 > 6 -10 {Combining like terms on either side}

4x > -4 {Dividing by 4 on both sides}

4x/4 > -4/4 {Simplifying each fraction}

x > - 1

Range of 2x − 5 < 7 and 4x + 10 > 6 will be all the x value which satisfy both the inequality x < 6 and x > -1.

Conclusion Part 1:

The range of all real numbers x such that 2x − 5 < 7 and 4x + 10 > 6 is can be given by interval notation (-1, 6)

Solution for part 2:

2x − 5 < 7 or 4x + 10 > 6 is called a compound inequality. To find the range we need to solve each of the inequality and range would be the values of x which The solution is same as part 1, with a small difference.

Range of 2x − 5 < 7 or 4x + 10 > 6 will be all the x value which satisfy any one of the inequality x < 6 or x > -1.

Conclusion Part 2:

The range of all real numbers x such that 2x − 5 < 7 or 4x + 10 > 6 is can be given by interval notation (-∞, ∞)

Note:

(i) When two inequalities are joined by the word and, the solution of the compound inequality occurs when both inequalities are true at the same time. It is the overlap, or intersection, of the solutions for each inequality.

(ii) When the two inequalities are joined by the word or, the solution of the compound inequality occurs when either of the inequalities is true. The solution is the combination, or union, of the two individual solutions.

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.

c. 1,233 miles