The greatest common factor of 42 & 15

Mathematics

1 answer:

Readme [11.4K]2 years ago

3 0

42 : 2 = 21

21 : 3 = 7

7 : 7 = 1

42 = 2 · 3 · 7

15 : 3 = 5

5 : 5 = 1

15 = 3 · 5

GCF(42, 15) = 3

You might be interested in

What formulas do I use to find surface area of cube, rectangular prism, trapezodial prism, triangular prism

-BARSIC- [3]

Cube: Area cubed
Rectangular Prism: A=(wl+hl+hw)

7 0

2 years ago

Simply (6x -2) 2 (0.5) 4

juin [17]

(6x -2) 2 (0.5) 4
(6x -2) 4
24x -8
Solution
24x -8

5 0

2 years ago

Calculus Problem

Roman55 [17]

The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$