All categories

3 years ago

What is 1386/1000 in simplest form

1 answer:

4 0

If you would like to write 1386/1000 in the simplest form, you can do this using the following steps:

1386/1000 = 693/500 = 1 193/500

The correct result would be 693/500.

You might be interested in

Determine the slope and the y-Intercept of the line that passes through A(-2,4), B(0,8)

Semenov [28]

Answer and work down below. Let me know if you have questions

8 0

3 years ago

A 3

Andrew [12]

Answer:

There are (63) combinations. The notation means "six choose three". Out of six items (flavors) choose three.

(nk)=n!k!(n−k)!.

(63)=6!3!3!.

Think of it this way. There are 6 ways to choose a flavor. Once you choose, there are 5 ways to choose the next. After that, there are 4 flavors left. which is 6!/3!=6⋅5⋅4⋅3⋅2⋅13⋅2⋅1=6⋅5⋅4=120.

But, you could have chosen {chocolate,vanilla,strawberry} and you get the same combination as {vanilla, strawberry, chocolate} so we have to divide by 3!=3⋅2⋅1=6 to account for the order of choosing.

So the number of combinations of flavors is (63)=1206=20.

<h3>Mark me a brainlist</h3>

7 0

3 years ago

Solve the inequality for x. Show each step of the solution. 16x > 12(2x – 6) – 24

Dafna11 [192]

You want to start this problem by handling the parentheses and distributing the 12.
So 16x > 12(2x - 6) - 24 would become...
So 16x > 24x - 72 - 24
Next you would combine like terms...
So 16x > 24x - 96
Then to get rid of the negative you would add 96 to both sides...
So 16x + 96 > 24x
After that you would subtract 16x from both sides...
96 > 8x
Lastly, divide by 8 to get x.
12 > x

4 0

4 years ago

❊ Simplify :

DiKsa [7]

Answer:

See Below.

Step-by-step explanation:

Problem 1)

We want to simplify:

$\displaystyle \frac{a+2}{a^2+a-2}+\frac{3}{a^2-1}$

First, let's factor the denominators of each term. For the second term, we can use the difference of two squares. Hence:

$\displaystyle =\frac{a+2}{(a+2)(a-1)}+\frac{3}{(a+1)(a-1)}$

Now, create a common denominator. To do this, we can multiply the first term by (a + 1) and the second term by (a + 2). Hence:

$\displaystyle =\frac{(a+2)(a+1)}{(a+2)(a-1)(a+1)}+\frac{3(a+2)}{(a+2)(a-1)(a+1)}$

Add the fractions:

$\displaystyle =\frac{(a+2)(a+1)+3(a+2)}{(a+2)(a-1)(a+1)}$

Factor:

$\displaystyle =\frac{(a+2)((a+1)+3)}{(a+2)(a-1)(a+1)}$

Simplify:

$\displaystyle =\frac{a+4}{(a-1)(a+1)}$

We can expand. Therefore:

$\displaystyle =\frac{a+4}{a^2-1}$

Problem 2)

We want to simplify:

$\displaystyle \frac{1}{(a-b)(b-c)}+\frac{1}{(c-b)(a-c)}$

Again, let's create a common denominator. First, let's factor out a negative from the second term:

$\displaystyle \begin{aligned} \displaystyle &= \frac{1}{(a-b)(b-c)}+\frac{1}{(-(b-c))(a-c)}\\\\&=\displaystyle \frac{1}{(a-b)(b-c)}-\frac{1}{(b-c)(a-c)}\\\end{aligned}$