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RoseWind [281]
3 years ago
8

-8x/5+1/6=-5x/3 I don't get how to find the answer?

Mathematics
1 answer:
NeX [460]3 years ago
3 0
In order to find this answer we have to manipulate the equation algebraically so that we end up with a statement that x = something. We can manipulate the problem however we need to as long as we do the same thing to both sides of the equation (this keeps the equation true).

-8x/5 + 1/6 = -5x/3

first lets get the x's on one side of the equals sign by adding 8x/5 to each side.

1/6 = -5x/3 + 8x/5

Ok, now let's add the x's together (we need common denominators)

1/6 = -25x/15 + 24x/15
1/6 = -x/15

Now lets get x by itself by multiplying each side by 15

1/6 * 15 = -x
15/6 = -x
3/2 = -x

Multiply each side by -1 to make the x positive.

-3/2 = x
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Answer:

1: 5/6

2: 3 1/6

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hope this helps!!

p.s

    comment if wrong

8 0
3 years ago
The height of a parallelogram is three times its base. If the area is 972 square inches, what is the base and height?
Fed [463]
So h is 3x 
b is x
3x plus x equal 972 
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6 0
3 years ago
Let alpha and beta be conjugate complex numbers such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}
miskamm [114]

Answer:

-3+i\sqrt{3} , 1+\sqrt{3}

Step-by-step explanation:

Given that alpha and beta be conjugate complex numbers

such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}.

Let

\alpha = x+iy\\\beta = x-iy

since they are conjugates

\alpha-\beta = x+iy-(x-iy)\\= 2iy= 2i\sqrt{3} \\y =\sqrt{3}

\frac{\alpha}{\beta^2} }\\=\frac{x+i\sqrt{3} }{(x-i\sqrt{3})^2} \\=\frac{x+i\sqrt{3}}{x^2-3-2i\sqrt{3}} \\=\frac{x+i\sqrt{3}((x^2-3+2i\sqrt{3}) }{(x^2-3-2i\sqrt{3)}(x^2-3-2i\sqrt{3})}

Imaginary part of the above =0

i.e. \sqrt{3} (x^2-3)+2x\sqrt{3} =0\\x^2+2x-3=0\\(x+3)(x-1) =0\\x=-3,1

So the value of alpha = -3+i\sqrt{3} , 1+\sqrt{3}

3 0
3 years ago
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