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Lapatulllka [165]
3 years ago
12

Take away 4/5 from 6 1/2​

Mathematics
2 answers:
Rashid [163]3 years ago
8 0

Hi there!  

»»————- ★ ————-««

I believe your answer is:  

5\frac{7}{10}

»»————- ★ ————-««  

Here’s why:  

⸻⸻⸻⸻

\boxed{\text{Calculating the answer...}}\\\\6\frac{1}{2}-\frac{4}{5} \\-------------\\6\frac{1}{2} = \frac{13}{2} \\\\\frac{13}{2} - \frac{4}{5}\\\\LCM(2,5): 10\\\\\frac{13}{2} =\frac{13*5}{2*5} =\frac{65}{10}\\\\\frac{4}{5}=\frac{4*2}{5*2}=\frac{8}{10}\\\\\frac{65}{10}-\frac{8}{10}\\\\  \frac{57}{10}\\\\\frac{57}{10}\rightarrow\boxed{5\frac{7}{10}}

»»————- ★ ————-««  

Hope this helps you. I apologize if it’s incorrect.  

Dmitry [639]3 years ago
6 0

Answer:

3-4/5=2.2

hope it helps

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He determines that 6 is an extraneous solution because the difference of the numerators is 6, so the 6s cancel to 0. which best
seropon [69]

The question is incomplete. Completed question is given below the answer.

His solution for x is correct, but in order for 6 to be an extraneous solution, one denominator has to result in 0 when 6 is substituted for x.

Given  

Then the solution follows thus:

Step 1: 8(x – 4) = 2(x + 2)

Step 2: 4(x – 4) = (x + 2)

Step 3: 4x – 16 = x + 2

Step 4: 3x = 18

Step 5: x = 6

It can be seen that his solution is correct. But 6 is not an extraneous solution.

An extraneous solution is a solution to an equation that emerges from the process of solving the problem but is not a valid solution to the original problem.

When 6 is substituted into the original equation, the original equation holds.

Therefore, his solution for x is correct, but in order for 6 to be an extraneous solution, one denominator has to result in 0 when 6 is substituted for x.

Learn more about extraneous solution here: brainly.com/question/3751209

#SPJ4

Completed question:-

A student solves the following equation for all possible values of x:His solution is as follows:

Step 1: 8(x – 4) = 2(x + 2)

Step 2: 4(x – 4) = (x + 2)

Step 3: 4x – 16 = x + 2

Step 4: 3x = 18

Step 5: x = 6

He determines that 6 is an extraneous solution because the difference of the numerators is 6, so the 6s cancel to 0.

Which best describes the reasonableness of the student’s solution?

His solution for x is correct and his explanation of the extraneous solution is reasonable.

His solution for x is correct, but in order for 6 to be an extraneous solution, both denominators have to result in 0 when 6 is substituted for x.

His solution for x is correct, but in order for 6 to be an extraneous solution, one denominator has to result in 0 when 6 is substituted for x.

His solution for x is incorrect. When solved correctly, there are no extraneous solutions.

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Answer:

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