What is the GCF of 6,42, and 12

Answers

Answer 1
Answer: The GCF of 6, 42, and 12 is: 

6. 

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Which is true about the degree of the sum and difference of the polynomials 3xy - 2xy4 - 7xy and -8xy + 2x°y4 + xy?O Both the sum and difference have a degree of 6.
O Both the sum and difference have a degree of 7.
The sum has a degree of 6, but the difference has a degree of 7.
The sum has a degree of 7, but the difference has a degree of 6.

Answers

The degree of the sum and difference of the polynomials are 6 and 7 respectively.

The degree of the sum and difference of the polynomials.

3x⁵y - 2x³y⁴ - 7xy³ and -8x⁵y + 2x³y⁴ + xy³

The degree of sum = monomial with the highest power = 5 + 1 = 6

The degree of difference = monomial with the highest power = 3 + 4 = 7

Therefore, the degree of the sum and difference of the polynomials are 6 and 7 respectively.

To get more about polynomials refer to:

brainly.com/question/1600696

#SPJ7

Answer:

It's C I just did it

Step-by-step explanation:

If anyone can help me answer this it would be great thx

Answers

ok so remember
area=pir^2
area=256pi

256pi=pir^2
divide both sides by pi
256=r^2
squaer root
16=r

find r and d
d=2r
d=2(16)
d=32

radius=16ft
diameter=32ft




RADIUS:A=3.14*r^2
area=256pi

256*3.14=3.14*r^2
divide both sides by 3.14
256=r^2
square root
16=r


DIAMETER:
3.Now double the radius to get the diameter.32 is the diameter.


radius=16ft
diameter=32ft

Please Help What is the perimeter of a rhombus with diagonals that have a length of 12 and 9?I need application of rhombus properties and definitions, calculations, and application of perimeter.

Answers

Answer:

12×2+9×2=Perimeter.This a very easy

Amber is using 3 ft by14 ft sheets or wallpaper to cover a rectangular wall that has an area of 240ft^2. What is the least number of sheets of wallpaper she will need?

Answers

3ft x 14ft = 42ft^2


 240ft^2
----------
 42ft^2

5.7

So she needs 6 sheets. 

What's 3,897.003 in expanded form.

Answers

3,897.003 in expanded form is:

3,000 + 800 + 90 + 7 + 0.003

(Essay Question 7 Points)Suppose you roll a six-sided die two times hoping to get two numbers whose sum is even. What is the sample space? How many favorable outcomes are there?

Answers


Well I don't know.
Let's think about it:

-- There are 6 possibilities for each role.
    So 36 possibilities for 2 rolls.
    Doesn't take us anywhere.

New direction:
-- If the first roll is odd, then you need another odd on the second one.
-- If the first roll is even, then you need another even on the second one.
This may be the key, right here !

-- The die has 3 odds and 3 evens.

-- Probability of an odd followed by another odd = (1/2) x (1/2) = 1/4
-- Probability of an even followed by another even = (1/2) x (1/2) = 1/4

I'm sure this is it.  I'm a little shaky on how to combine those 2 probs.

Ah hah ! 
Try this:

Probability of either 1 sequence or the other one is (1/4) + (1/4) = 1/2 .

That means ... Regardless of what the first roll is, the probability of
the second roll matching it in oddness or evenness is 1/2 .

So the probability of 2 rolls that sum to an even number is 1/2 = 50% .

Is this reasonable, or sleazy ?