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lozanna [386]
3 years ago
13

6-5 solving rational equations 8/x+3=1/x+1

Mathematics
1 answer:
AfilCa [17]3 years ago
7 0

Answer:

x = -5/7

Step-by-step explanation:

8/(x+3)=1/(x+1)

Using cross products

8*(x+1) = 1*(x+3)

Distribute

8x+8 = x+3

Subtract x from each side

8x-x +8 = x+3-x

7x +8 =3

Subtract 8 from each side

7x+8-8 = 3-8

7x = -5

Divide by 7

7x/7 = -5/7

x = -5/7

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The curves r1(t) = 2t, t2, t4 and r2(t) = sin t, sin 5t, 2t intersect at the origin. Find their angle of intersection, θ, correc
masya89 [10]

Answer:

Therefore the angle of intersection is \theta =79.48^\circ

Step-by-step explanation:

Angle at the intersection point of two carve is the angle of the tangents at that point.

Given,

r_1(t)=(2t,t^2,t^4)

and r_2(t)=(sin t , sin5t, 2t)

To find the tangent of a carve , we have to differentiate the carve.

r'_1(t)=(2,2t,4t^3)

The tangent at (0,0,0) is     [ since the intersection point is (0,0,0)]

r'_1(0)=(2,0,0)      [ putting t= 0]

|r'_1(0)|=\sqrt{2^2+0^2+0^2} =2

Again,

r'_2(t)=(cos t ,5 cos5t, 2)

The tangent at (0,0,0) is    

r'_2(0)=(1 ,5, 2)        [ putting t= 0]

|r'_1(0)|=\sqrt{1^2+5^2+2^2} =\sqrt{30}

If θ is angle between tangent, then

cos \theta =\frac{r'_1(0).r'_2(0)}{|r'_1(0)|.|r'_2(0)|}

\Rightarrow cos \theta =\frac{(2,0,0).(1,5,2)}{2.\sqrt{30} }

\Rightarrow cos \theta =\frac{2}{2\sqrt{30} }

\Rightarrow cos \theta =\frac{1}{\sqrt{30} }

\Rightarrow  \theta =cos^{-1}\frac{1}{\sqrt{30} }

\Rightarrow  \theta =79.48^\circ

Therefore the angle of intersection is \theta =79.48^\circ.

8 0
3 years ago
F(x) = x°-9x
laila [671]
F(x)=x^3-9x
and
g(x)=x^2-2x-3

so you just need to divide f(x) by g(x)

Therefore:

f(x)/g(x) = (x^3-9x) / (x^2-2x-3)

and of course you need to factor these two function to see if some factor would cancel another

x^3-9x = x(x^2-9)=x(x-3)(x+3)
and
x^2-2x-3 = (x-3)(x+1)





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