I don't know how to simplify

Mathematics

1 answer:

In-s [12.5K]2 years ago

6 0

(5-2/x)/4-3/x^2

after simplifying these
(5x-2)/x(4x^2-3)/x^2
x(5x-2)/(4x^2-3)
5x^2-2x/4x^2-3

now u can solve it

You might be interested in

Can someone please help me out? It would really make me happier if you did.

asambeis [7]

Answer:

B) 8 and 24.

Step-by-step explanation:

Just by looking at the answer choices, since one number must be three times the other, we can determine that the correct answer is B.

Regardless, let's work this out mathematically. Let the first number be a and the second number be b.

Their difference is 16. Hence:

$a-b=16$

The first number is three times the second number. So:

$a=3b$

Substitute:

$3b-b=16$

Solve for b. Subtract:

$2b=16$

And divide both sides by two:

$b=8$

So, the second number is 8.

And since the first number is three times the second, the first number is 24.

Our answer is B as expected.

5 0

2 years ago

A bag contains different colored candies. There are 50 candies in the bag, 28 are red, 10 are blue, 8 are green and 4 are yellow

Mrrafil [7]

Answer:

$\displaystyle \frac{54}{5405}$ .

Step-by-step explanation:

How many unique combinations are possible in total?

This question takes 5 objects randomly out of a bag of 50 objects. The order in which these objects come out doesn't matter. Therefore, the number of unique choices possible will the sames as the combination

$\displaystyle \left(50\atop 5\right) = 2,118,760$ .

How many out of that 2,118,760 combinations will satisfy the request?

Number of ways to choose 2 red candies out a batch of 28:

$\displaystyle \left( 28\atop 2\right) = 378$ .

Number of ways to choose 3 green candies out of a batch of 8:

$\displaystyle \left(8\atop 3\right)=56$ .

However, choosing two red candies out of a batch of 28 red candies does not influence the number of ways of choosing three green candies out of a batch of 8 green candies. The number of ways of choosing 2 red candies and 3 green candies will be the product of the two numbers of ways of choosing

$\displaystyle \left( 28\atop 2\right) \cdot \left(8\atop 3\right) = 378\times 56 = 21,168$ .