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3 years ago

Given any two numbers which is greater the least common multiple of the numbers or the greatest common factors of the numbers

Mathematics

1 answer:

tamaranim1 [39]3 years ago

8 0

THE LEAST COMMON MULTIPLE WILL BE HIGHER THAN THE greatest common factor as long as the 2 numbers are different.

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A graphed line with a slope of -1/3 and passing through a point of (3,-4)

LenaWriter [7]

Answer:

$y=-\frac{1}{3}x-3$

Step-by-step explanation:

we know that

The equation of the line in point slope form is equal to

$y-y1=m(x-x1)$

we have

$m=-\frac{1}{3}$

$(x1,y1)=(3,-4)$

substitute

$y+4=-\frac{1}{3}(x-3)$

Convert to slope intercept form

isolate the variable y

$y+4=-\frac{1}{3}x+1$

$y=-\frac{1}{3}x+1-4\\y=-\frac{1}{3}x-3$

3 0

4 years ago

Work out the percentage to change to decimal places when a price of 200 is increased to 210.99

olya-2409 [2.1K]

Answer:

105.495%

Step-by-step explanation:

210.99/200=1.05495

200 * 1.05495 = 210.99

6 0

3 years ago

George Box, a famous statistician, once said, "All models are wrong, but some are useful.” What did George Box mean by this stat

Katena32 [7]

Basically meaning that no model can exactly be on point with whatever it is modeling. Two things cannot be EXACTLY identical to what they are referring to, but it is saying that some can get close enough to be useful as an example.

3 0

3 years ago

Read 2 more answers

Find the length of the diagonal BD in the quadrilateral ABCD shown in the coordinate plane ?

Leno4ka [110]

bearing in mind B is at (4,3) and D is at (-2,-4).

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{4}~,~\stackrel{y_1}{3})\qquad D(\stackrel{x_2}{-2}~,~\stackrel{y_2}{-4})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ BD=\sqrt{(-2-4)^2+(-4-3)^2}\implies BD=\sqrt{(-6)^2+(-7)^2} \\\\\\ BD=\sqrt{36+49}\implies BD=\sqrt{85}$

6 0

3 years ago

Read 2 more answers

If x^2=20 what is the value of x will give brainliest for answer

Schach [20]

Answer:

x² - 20 = 0

Using the quadratic formula

$x = \frac{ - b± \sqrt{( {b})^{2} - 4ac } }{2a}$

a = 1 b = 0 c = -20

So we have

$x = \frac{ - 0 ± \sqrt{ {0}^{2} - 4(1)( -20)} }{2(1)} \\ \\ x = \frac{± \sqrt{80} }{2} \\ \\ x = \frac{±4 \sqrt{5} }{2} \\ \\ \\ x = ±2 \sqrt{5} \\ \\ \\ x = 2 \sqrt{5} \: \: \: or \: \: \: x = - 2 \sqrt{5}$

Hope this helps you.

8 0

3 years ago