Johanna used tiles to build a rectangular array with an area of 54. list all the possible dimension of the arrey

1 answer:

4 0

Really this is an exercise in factorization. First find all the factors of 54: 1, 2, 3, 6, 9, 18, 27, 54

Then use each of those in a multiplication to get 54:

1×54
2×27
3×18
6x9

That's it. 4 possibilities.

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A 12" diameter ball rolls around inside the bottom of a rectanglar bow that is 14" wide and 30" long.The ball always touches the

julsineya [31]

Answer:

1. Shape of line drawn by round ball is: Rectangle

2. Length of path round ball creates is: 64 inches

Step-by-step explanation:

The perimeter of the box is 30+30+14+14 = 88 inches.

We find a rectangle path of the round ball to be interior rectangle shape and measures 64 inches.

The shape of the path is rectangle, this is because when the balls direction from horizontal to vertical is drawn from the central line to the ball - the ball still rotates but such central line stays stationary in a straight line at same distance from edge as it changes direction so therefore it stays perpendicular and creates a rectangle shape. Only an oval ball would draw a line of oval shape due to its central line curving against the perpendicular edge of the rectangle. The round ball cannot create a square shape as the sides of the rectangle when reduced by 6'' + 6'' either side of length 30 = 18 inches and is still longer than the shorter sides of the box being 14'' which becomes 14- 6=8 inches to each side.

2. To find the path length- we look for the difference in width of the central ball first which is 12'' divided by 2 = 12/2 = 6 = 6 inches. We can then deduct this from 14'' and from 30'' then multiply by 2.

Width of one side = 14-6 = 8

= 8 inches

Length of one side = 30-6 = 24

= 24 inches

We add together 8+24 = 32, and then multiply by 2

32 *2 = 64

3 0

3 years ago

For the function f(x)=2x^3-3x^2-22.5x+19

kvv77 [185]

A. $f(x)=2x^3-3x^2-22.5x+19\\f'(x)=6x^2-6x-22.5\\f'(x)=6x^2-6x-22.5=0\\4x^2-4x-15=0\\(2x+3)(2x-5)=0\\2x+3=0~|~2x-5=0\\2x=-3~|~2x=5\\x=-1.5~|~x=2.5\\x=-1.5,2.5\\\\f''(x)=12x-6\\12x-6=0\\12x=6\\x=0.5$ - Relative maximum at x = -1.5. Relative minimum at x = 2.5. Point of inflection at x = 0.5.

B. $f(x)=2x^3-3x^2-22.5x+19\\f'(x)=6x^2-6x-22.5\\f'(x)=6x^2-6x-22.5=0\\4x^2-4x-15=0\\(2x+3)(2x-5)=0\\2x+3=0~|~2x-5=0\\2x=-3~|~2x=5\\x=-1.5~|~x=2.5\\x=-1.5,2.5$ - Absolute maximum at x = -1.5. Absolute minimum at x = 2.5. This applies for the interval [-4, 5].

C. $f(x)=2x^3-3x^2-22.5x+19\\f(0)=2(0)^3-3(0)^2-22.5(0)+19\\f(0)=19\\\\f'(x)=6x^2-6x-22.5\\f'(x)=6x^2-6x-22.5=0\\4x^2-4x-15=0\\(2x+3)(2x-5)=0\\2x+3=0~|~2x-5=0\\2x=-3~|~2x=5\\x=-1.5~|~x=2.5\\x=2.5$ - Absolute maximum at x = 0. Absolute minimum at x = 2.5. This applies for the interval [0, 3].

D. $f(x)=2x^3-3x^2-22.5x+19\\f'(x)=6x^2-6x-22.5\\f'(x)=6x^2-6x-22.5=0\\4x^2-4x-15=0\\(2x+3)(2x-5)=0\\2x+3=0~|~2x-5=0\\2x=-3~|~2x=5\\x=-1.5~|~x=2.5\\x=-1.5,2.5$ - Test: $f(-2)=2x^3-3x^2-22.5x+19\\f(-2)=2(-2)^3-3(-2)^2-22.5(-2)+19\\f(-2)=2(-8)-3(4)-45+19\\f(-2)=-16-12-26\\f(-2)=-540\\\\f(3)=2x^3-3x^2-22.5x+19\\f(3)=2(3)^3-3(3)^2-22.5(3)+19\\f(3)=2(27)-3(9)-67.5+19\\f(3)=54-27-48.5\\f(3)=-21.5$ - Increasing: $[-\infty,-1.5)(2.5,\infty]$ - Decreasing: $(-1.5,2.5)$

E. $f(x)=2x^3-3x^2-22.5x+19\\f'(x)=6x^2-6x-22.5\\f'(x)=6x^2-6x-22.5=0\\4x^2-4x-15=0\\(2x+3)(2x-5)=0\\2x+3=0~|~2x-5=0\\2x=-3~|~2x=5\\x=-1.5~|~x=2.5\\x=-1.5,2.5\\\\f''(x)=12x-6\\12x-6=0\\12x=6\\x=0.5$ - Concave up: $(0.5,\infty]$ - Concave down: $[-\infty,0.5)$