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3 years ago

What is the least common multiple of 4 and 2?

Mathematics

1 answer:

lina2011 [118]3 years ago

8 0

4 is the LCM because 2 times 1 is 2 and 2 times 2 is 4 << and going to 4 times 1 is 4 << so thats why ( 4 ) is the LCM

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Brighten is an expression that shows how to multiply 7 * 256 using expanded form and distributive property

charle [14.2K]

I am not familiar with "brighten" but this is how you can expand and use the distributive property:

7(256)

= 7(200 + 50 + 6)

= 7(200) + 7(50) + 7(6)

= 1400 + 350 + 30

= 1780

4 0

3 years ago

What's is 5x2/3 <br> A. 3 1/3 <br> B 3 2/5 <br> C 5 2/3<br> D 10/15

Sophie [7]

Answer:

Step-by-step explanation:

5 x 2/3=3.333333333...

and 3.3333333...

is also equal to 3 1/3

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3 years ago

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Lorico [155]

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8 0

4 years ago

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Match each interval with its corresponding average rate of change for q(x) = (x + 3)2. 1. -6 ≤ x ≤ -4 1 2. -3 ≤ x ≤ 0 -4 3. -6 ≤

MrMuchimi

The average rate of change of a function f(x) in an interval, a < x < b is given by
$\frac{f(b) - f(a)}{b - a}$

Given q(x) = (x + 3)^2

1.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -4 is given by $\frac{q(-4)-q(-6)}{-4-(-6)} = \frac{(-4+3)^2-(-6+3)^2}{-4+6} = \frac{1-9}{2} = \frac{-8}{2} =-4$

2.) The average rate of change of q(x) in the interval -3 ≤ x ≤ 0 is given by $\frac{q(0)-q(-3)}{0-(-3)} = \frac{(0+3)^2-(-3+3)^2}{0+3} = \frac{9-0}{3} = \frac{9}{3} =3$

3.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -3 is given by $\frac{q(-3)-q(-6)}{-3-(-6)} = \frac{(-3+3)^2-(-6+3)^2}{-3+6} = \frac{0-9}{3} = \frac{-9}{3} =-3$

4.) The average rate of change of q(x) in the interval -3 ≤ x ≤ -2 is given by $\frac{q(-2)-q(-3)}{-2-(-3)} = \frac{(-2+3)^2-(-3+3)^2}{-2+3} = \frac{1-0}{1} = \frac{1}{1} =1$

5.) The average rate of change of q(x) in the interval -4 ≤ x ≤ -3 is given by $\frac{q(-3)-q(-4)}{-3-(-4)} = \frac{(-3+3)^2-(-4+3)^2}{-3+4} = \frac{0-1}{1} = \frac{-1}{1} =-1$

6.) The average rate of change of q(x) in the interval -6 ≤ x ≤ 0 is given by $\frac{q(0)-q(-6)}{0-(-6)} = \frac{(0+3)^2-(-6+3)^2}{0+6} = \frac{9-9}{6} = \frac{0}{6} =0$

3 0

3 years ago

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I need the help pls

Gnoma [55]

Answer:

A,C,B

is the answer

3 0

3 years ago