All categories

3 years ago

PLEASE HELP ME me me me

Mathematics

2 answers:

kykrilka [37]3 years ago

7 0

Answer:

Andre has the correct answer. When simplified his answer is equivalent to the original equation.

allsm [11]3 years ago

6 0

Answer:

2pg1

Step-by-step explanation:

Hehe

You might be interested in

Please help i will mark brainliest!

damaskus [11]

Answer:

4,2,1

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Solve for r in terms of q, s, t and u. <br> St= -ruq

Anni [7]

Answer:

hope this helps

give brainliest

4 0

3 years ago

Write a for loop that computes the following sum: 5 10 15 20 ... 485 490 495 500. The sum should be placed in a variable sum tha

Varvara68 [4.7K]

The solution is in the attachment

8 0

3 years ago

A triangle has vertices (2, 1), (5, 1), and (5,5). What is the length of the longest side of the triangle?

Lerok [7]

Answer:

5,5

Step-by-step explanation:

it has the longest side

8 0

2 years ago

Which statement best reflects the solution(s) of the equation?

Inessa [10]

x=2 is only solution while x=1 is extraneous solution

Option C is correct.

Step-by-step explanation:

We need to solve the equation $\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}$ and find values of x.

Solving:

Find the LCM of denominators x-1,x and x-1. The LCM is x(x-1)

Multiply the entire equation with x(x-1)

$\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}\\\frac{1}{x-1}*x(x-1)+\frac{2}{x}*x(x-1)=\frac{x}{x-1}*x(x-1)\\Cancelling\,\,out\,\,the\,\,same\,\,terms:\\x+2(x-1)=x^2\\x+2x-2=x^2\\3x-2=x^2\\x^2-3x+2=0$

Now, factoring the term:

$x^2-2x-x+2=0\\x(x-2)-1(x-2)=0\\(x-1)(x-2)=0\\x-1=0\,\,and\,\, x-2=0\\x=1\,\,and\,\, x=2$

The values of x are x=1 and x=2

Checking for extraneous roots:

Extraneous roots: The root that is the solution of the equation but when we put it in the equation the answer turns out not to be right.

If we put x=1 in the equation, $\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}$ the denominator becomes zero i.e

$\frac{1}{1-1}+\frac{2}{1}=\frac{1}{1-1}\\\frac{1}{0}+2=\frac{1}{0}$

which is not correct as in fraction anything divided by zero is undefined. So, x=1 is an extraneous solution.

If we put x=2 in the equation,

$\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}$

$\frac{1}{2-1}+\frac{2}{2}=\frac{2}{2-1}\\\frac{1}{1}+1=\frac{2}{1}\\1+1=2\\2=2$

So, x=2 is only solution while x=1 is extraneous solution

Option C is correct.

Keywords: Solving Equations and checking extraneous solution

Learn more about Solving Equations and checking extraneous solution at:

#learnwithBrainly

5 0

3 years ago

PLEASE HELP ME me me me​

PLEASE HELP ME me me me