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4 years ago

10

(6/7^c)^2 Please Help

1 answer:

Aleksandr [31]4 years ago

5 0

36/7^2c that’s what I got I’m sure that is right

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

Dmitriy789 [7]

Answer:

Solution: $x=-\frac{1}{5}$ , $x=0$

Step-by-step explanation:

Given:

The rational equation to solve is given as:

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

Doing cross product, we get:

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

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.

So, no extraneous solution.

4 0

4 years ago

How many feet are in 1 yard? i yard is how many feet

Katena32 [7]

There are 3 feet in one yard

3 0

3 years ago

Read 2 more answers

Need help on this too

andrezito [222]

Answer:

I think it is distributive

Step-by-step explanation:

When a single number is next to parentheses, it's called distributive property

I hope this is right :/

Please mark brainliest :/

6 0

3 years ago

Use properties to rewrite the given equation. Which equations have the same solution as 3/5x +2/3 + x = 1/2– 1/5x? Check all tha

vodomira [7]

we have

$\frac{3}{5}x+ \frac{2}{3}+x=\frac{1}{2}-\frac{1}{5}x$

Combine like terms in both sides

$(\frac{3}{5}x+ x)+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$

we know that

$(\frac{3}{5}x+ x)=(\frac{3}{5}x+ \frac{5}{5}x)=\frac{8}{5}x$

substitute in the expression above

$\frac{8}{5}x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$ -----> equation A

Multiply equation A by $5*3*2=30$ both sides

$30*(\frac{8}{5}x+\frac{2}{3})=30*(\frac{1}{2}-\frac{1}{5}x)$

$48x+20=15-6x$ ---------> equation B

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$48x+6x=15-20$

$54x=-5$ ---------> equation C

Solve for x

$x=-\frac{5}{54} =-0.09$

We are going to proceed to verify each case to determine the solution.

<u>Case a)</u> $\frac{8}{5}x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$

the case a) is equal to the equation A

so

the case a) have the same solution that the given equation

<u>Case b)</u> $18x+20+30x=15-6x$

Combine like terms in left side

$(18x+30x)+20=15-6x$

$(48x)+20=15-6x$

the case b) is equal to the equation B

so

the case b) have the same solution that the given equation

<u>Case c)</u> $18x+20+x=15-6x$

Combine like terms in left side

$(18x+x)+20=15-6x$

$(19x)+20=15-6x$

$19x+6x=15-20\\25x=-5\\x=-0.20$

$-0.20\neq -0.09$

therefore

the case c) not have the same solution that the given equation

<u>Case d)</u> $24x+30x=-5$

Combine like terms in left side

$54x=-5$

the case d) is equal to the equation C

so

the case d) have the same solution that the given equation

<u>Case e)</u> $12x+30x=-5$

Combine like terms in left side

$42x=-5$

$x=-5/42=-0.12$

$-0.12\neq -0.09$

therefore

the case e) not have the same solution that the given equation

therefore

<u>the answer is</u>

case a) $\frac{8}{5}x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$

case b) $18x+20+30x=15-6x$

case d) $24x+30x=-5$

7 0

4 years ago

Read 2 more answers

Anyone know how to factor 4x^2-4x-80?

andrezito [222]

(x-5)(4x+16) Is the answer
ASK ME IF YOU GOT A QUESTION, I'M GLAD TO HELP. ;)

8 0

4 years ago