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4 years ago

How can 6691 divided by 28 66 can go into 3?

2 answers:

7 0

If you want to know wheather 3 can go into a number,heres a trick add up the Numbers tge sum is divisible by 3 then the number itself is divisible by 3

AlladinOne [14]4 years ago

3 0

If you want to know weather 3 can go into a number add up the Numbers the sum is divisible by 3 then the number itself is divisible by 3

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What calculation can I use to find 48 after a 49% decrease?

Natasha_Volkova [10]

Answer:

Step-by-step explanation:

hope it helps

7 0

3 years ago

Will mark branliest combine like terms to create an equivalent expression to 7(2j + 3 + 2j).

Helga [31]

Answer:

28j + 21

step-by-step explanation:

(7) (2j+3+2j)

=(7*2j) + (7*3) + (7*2j)

=14j + 21 + 14j

=28j + 21

7 0

3 years ago

Read 2 more answers

PLEASE PLEASE I RLY NEED HELP ...... ....... . .. . . . . . . . . .... . . . Pick three Enlightenment thinkers and explain how t

just olya [345]

Answer:

Enlightenment philosophers John Locke, Charles Montesquieu, and Jean-Jacques Rousseau all developed theories of government in which some or even all the people would govern. These thinkers had a profound effect on the American and French revolutions and the democratic governments that they produced.

I am guessing your in the same situation as me sweeping across the internet for help in our summer school history classes well I found this hope it helps

3 0

3 years ago

Solve for x. 18 ÷ x = -2 x = _____

Eddi Din [679]

Answer:

-9

Step-by-step explanation:

18÷x = -2

18 = -2×x

18 = -2x

-9 = x

x = -9

6 0

3 years ago

Read 2 more answers

Solve the rational equation x divided by 6 equals x squared divided by quantity x minus 1 end quantity, and check for extraneous

pychu [463]

Answer:

$x = 0$ and $x = \textrm{negative one fifth}$

Step-by-step explanation:

Given:

The rational equation to solve is given as:

$\frac{x}{6}=\frac{x^2}{x-1}
$

We use the cross multiplication rule and multiply the numerators one side with the denominators of the other side. Therefore,

$x\times (x-1)=6\times x^2\\x^2-x=6x^2\\\textrm{Bringing all variables to the right side, we get:}\\6x^2-x^2+x=0\\5x^2+x=0\\x(5x+1)=0\\x=0\ or\ 5x+1=0\\x=0\ or\ x=-\frac{1}{5}
$

Now, for $x=0$ , the rational equation is equal to 0. So, $x=0$ is a solution.

Also, $x=-\frac{1}{5}$ is also a solution as the equation's denominator is defined at that point.

So, no extraneous solution.

8 0

4 years ago