Question

PLEASE HELP ASAP!!! WILL MARK BRAINLIEST!!!!
Put a positive factor back into the square root:
$\sqrt[-5]{0.6}$
Solve for x
(2x-a)/b=(ax+1)/c, if ab ≠ 2c

Answer 1

Answer:

Part 1) $-\sqrt{15}$

Part 2) $x=\frac{b+ac}{2c-ab}$

Step-by-step explanation:

Part 1) we know that

To put factor back into the square root, we have to put squared value

we have

$-5\sqrt{0.6}$

Remember that

$5=\sqrt{5^2}$

substitute in the expression above

$-5\sqrt{0.6}=-\sqrt{(5^2)(0.6)}=-\sqrt{25*0.6}=-\sqrt{15}$

Part 2) we have

$\frac{2x-a}{b}=\frac{ax+1}{c}$

Solve for x

That means----> Isolate the variable x

Multiply in cross

$(2x-a)c=(ax+1)b$

Apply distributive property

$2cx-ac=abx+b$

Group terms

$2cx-abx=b+ac$

Factor x left side

$x(2c-ab)=b+ac$

Divide by (2c-ab) both sides

$x=\frac{b+ac}{2c-ab}$