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dimaraw [331]
3 years ago
15

How do I solve this

Mathematics
1 answer:
anastassius [24]3 years ago
7 0
\bf \cfrac{x+2}{x^2+6x-7}    so, that function is "defined", ok, what values of "x" are not in the domain, namely, what values can "x" take on and not make the function "undefined", well,  you know, if we end up with a 0 at the denominator, like   \bf \cfrac{x+2}{0}    then, we'd have an "undefined" expression...so... any values of "x" that make the denominator 0, are not really the ones we want, and thus they'd be excluded from the domain.


so, hmm which are those? let's check, let's set the denominator to 0, and solve for "x".

\bf x^2+6x-7=0\implies (x+7)(x-1)=0\implies x=
\begin{cases}
-7\\
1
\end{cases}
\\\\\\
\textit{let's check, } x=-7\quad \cfrac{(-7)+2}{(-7)^2+6(-7)-7}\implies \cfrac{-5}{49-42-7}\implies \cfrac{-5}{0}
\\\\\\
x=1\quad \cfrac{(1)+2}{(1)^2+6(1)-7}\implies \cfrac{3}{1+6-7}\implies \cfrac{-3}{0}
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3 years ago
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Slove for x 5(2x-1)=6
german

Answer:

x = 11/10

Step-by-step explanation:

5(2x-1)=6

Distribute

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Add 5 to each side

10x-5+5 = 6+5

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Divide by 10

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The curves r1(t) = 4t, t2, t3 and r2(t) = sin(t), sin(5t), 2t intersect at the origin. find their angle of intersection, θ, corr
san4es73 [151]
First get the tangent vectors by differentiating r1 and r2 
r_1 '(t) = (4,2t,3t^2) \\  \\  r_2'(t) = (cos (t), 5 cos(5t), 2)
Evaluate at t=0
r_1' = (4,0,0) \\  \\ r_2' = (1,5,2)

Use identity for angle between 2 vectors:
u*v = |u| |v| cos \theta

Evaluate dot product and unit vectors:
u*v = (4,0,0)*(1,5,2) = 4 \\  \\ |u| = \sqrt{4^2 +0^2+0^2} = 4 \\  \\ |v| = \sqrt{1^2 + 5^2+2^2} = \sqrt{30}

Sub into identity and solve for theta:
4 = 4 \sqrt{30} cos \theta \\  \\ cos\theta = \frac{1}{\sqrt{30}} \\  \\ \theta = 79.48

Answer:
Angle of intersection is about 79 degrees.




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3 years ago
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