HELP ME QUICK PLEASEEEE solve for x: 2/x-2+7/x^2-4=5/x

Mathematics

1 answer:

Anna35 [415]3 years ago

3 0

$\displaystyle\\\\\frac{2}{x-2}+\frac{7}{x^2-4}=\frac{5}{x}\\\\\frac{^{x+2)}2~~~~~}{x-2}+\frac{7}{(x-2)(x+2)}=\frac{5}{x}\\\\\frac{2(x+2)}{(x-2)(x+2)}+\frac{7}{(x-2)(x+2)}=\frac{5}{x}\\\\\frac{2x+4+7}{x^2-4}=\frac{5}{x}\\\\\frac{2x+11}{x^2-4}=\frac{5}{x}\\\\5(x^2-4)=x(2x+11)\\\\5x^2-20=2x^2+11x\\\\5x^2-2x^2 -11x-20=0$

$\displaystyle\\3x^2-11x-20=0\\\\x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{11\pm\sqrt{121+4\cdot3\cdot20}}{2\cdot3}=\\\\=\frac{11\pm\sqrt{121+240}}{6}=\frac{11\pm\sqrt{361 }}{6}=\frac{11\pm19}{6}\\\\x_1=\frac{11-19}{6}=\frac{-8}{6}\\\\\boxed{\bf x_1=-\frac{4}{3}}\\\\x_2=\frac{11+19}{6}=\frac{30}{6}\\\\\boxed{\bf x_2=5}$

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A rhombus has four 6-inch sides and two 120-degree angles. From one of the vertices of the obtuse angles, the two latitudes are

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Answer:

Area(A)=Area(C)= $9 in^{2}$

Area(B)= $13.2 in^{2}$

Step-by-step explanation:

We begin with finding the angles a and b that from the drawing attached you can see that a=b.

Now, the sum of the internal angles of a rhomboid is equal to 360 degrees, with that we have:

120+120+a+b=360

240+2a=360

2a=120

a=60=b

Next, in the image you can see that the lines coming from the angle at the top 120 degrees vertex, divide the opposite sides by half, thus making two triangles with one side of 6 in and another of 3 in.

We can say from the drawing as well:

Area(A)+Area(B)+Area(C)=Area(rhomboid)

But, we can also say that Area(A)=Area(C)

So, starting with Area(A)

Area(A)=Area(triangle)= $\frac{b*h}{2}$ = $\frac{6*3}{2}$ = $9 in^{2}$

We can then calculate the area B, a rhomboid, or better, take the Total area of the figure and subtract the area of the two triangles.

Area(B)=Area(rhomboid)-Area(A)-Area(C)

Area(rhomboid)=b*h where b=6in and h is the perpendicular distance from the base to the top.

$h=[tex]6*cos(30)=5.20in$ The 30 degrees come from: 120-30-60=30, since the latitudes split the 120 angle in two equal parts and one that is the half of the obtuse angle.