All categories

2 years ago

Find the greatest common factor of 8 and 24.

Mathematics

1 answer:

oksano4ka [1.4K]2 years ago

6 0

Answer:

Step-by-step explanation:

Factors of 8: 1, 2, 4, 8

Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24

They do share a few factors but the greatest one shared between the two is 8

You might be interested in

Work out the values of a to e please

baherus [9]

Answer:

Step-by-step explanation:

a=10 (4,0,2,4,3,1,2,3,1,4)

b=4 (7,5,6,9)

c=6 (13,13,13,10,13,12)

d= 5 (17,16,17,17,16)

e=6 (21,23,20,20,21,23)

8 0

2 years ago

Which strategy would you use to find 2+8? a) doubles plus 1. b)count on. c) doubles. d) doubles minus 1

Gekata [30.6K]

You would count on.

I hope i am right

8 0

3 years ago

A swimming pool is 50 meters long,15 meters wide,and 3 meters deep. What is the volume of the pool

steposvetlana [31]

Answer: 2,250 $m^{3}$

Step-by-step explanation:

Volume of the pool is given by : Length X breadth X height

V = 50 X 15 X 3

V = 2,250

5 0

2 years ago

HELP PLEASE !!! 4th time asking this :(( i really need help PLEASE !

harina [27]

Answer:

A: (-5,4)

B: (7,-2)

Step-by-step explanation:

It has to be in order start with the x-axis then the y-axis

4 0

2 years ago

Evaluate the integral of the quantity x divided by the quantity x to the fourth plus sixteen, dx . (2 points) one eighth times t

Anika [276]

Answer:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

Step-by-step explanation:

Given

$\int\limits {\frac{x}{x^4 + 16}} \, dx$

Required

Solve

Let

$u = \frac{x^2}{4}$

Differentiate

$du = 2 * \frac{x^{2-1}}{4}\ dx$

$du = 2 * \frac{x}{4}\ dx$

$du = \frac{x}{2}\ dx$

Make dx the subject

$dx = \frac{2}{x}\ du$

The given integral becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{x}{x^4 + 16}} \, * \frac{2}{x}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{1}{x^4 + 16}} \, * \frac{2}{1}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

Recall that: $u = \frac{x^2}{4}$

Make $x^2$ the subject

$x^2= 4u$

Square both sides

$x^4= (4u)^2$

$x^4= 16u^2$

Substitute $16u^2$ for $x^4$ in $\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16u^2 + 16}} \,\ du$

Simplify

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16}* \frac{1}{8u^2 + 8}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{2}{16}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

In standard integration

$\int\limits {\frac{1}{u^2 + 1}} \,\ du = arctan(u)$

So, the expression becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(u)$

Recall that: $u = \frac{x^2}{4}$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

4 0

2 years ago