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snow_tiger [21]
3 years ago
11

Please help with this question

Mathematics
1 answer:
astraxan [27]3 years ago
7 0

Answer:

x = -8/2

Step-by-step explanation:

To make the equation easier to work with, our first step will be to make all of our fractions have a common denominator. Both 2 and 4 are factors of 8, so that will be our common denominator.

Old Equation: 1/4x - 1/8 = 7/8 + 1/2x

New Equation (with common denominators): 2/8x - 1/8 = 7/8 + 4/8x

Now, we're going to begin to isolate the x variable. First, we're going to subtract 2/8x from both sides, eliminating the first variable term on one side completely.

2/8x - 1/8 = 7/8 + 4/8x

-2/8x                    -2/8x

__________________

-1/8 = 7/8 + 2/8x

We're one step closer to our x variable being isolated. Next, we're going to move the constants to the left side of the equation. To do this, we must subtract by 7/8 on both sides.

-1/8 = 7/8 + 2/8x

- 7/8  -7/8

______________

-1 = 2/8x

Our last step is to multiply 2/8x by its reciprocal in order to get the x coefficient to be 1. This means multiply both sides by 8/2.

(8/2)  -1 = 2/8x (8/2)

The 2/8 and 8/2 cancel out, and you're left with:

-8/2 = x

I hope this helps!

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Check the picture below, now the distance from 2,0 to 4,0 there's no need to do much calculation since that's just 2 units, as you see there.

~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{2}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{4}) ~\hfill d1=\sqrt{[ 1- 2]^2 + [ 4- 0]^2} \\\\\\ d1=\sqrt{(-1)^2+4^2}\implies \boxed{d1=\sqrt{17}} \\\\[-0.35em] ~\dotfill

(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{-1}~,~\stackrel{y_2}{-2}) ~\hfill d2=\sqrt{[ -1- 1]^2 + [ -2- 4]^2} \\\\\\ d2=\sqrt{(-2)^2+(-6)^2}\implies \boxed{d2=\sqrt{40}} \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{-1}~,~\stackrel{y_1}{-2})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{0}) ~\hfill d3=\sqrt{[ 4- (-1)]^2 + [ 0- (-2)]^2} \\\\\\ d3=\sqrt{(4+1)^2+(0+2)^2}\implies \boxed{d3=\sqrt{29}}

~\dotfill\\\\ \stackrel{\textit{\Large Perimeter}}{2~~ + ~~\sqrt{17}~~ + ~~\sqrt{40}~~ + ~~\sqrt{29}~~\approx~~17.83}

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