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Tcecarenko [31]
4 years ago
5

X-7y=-21 2x - 14y = -42 What’s the solution

Mathematics
1 answer:
maxonik [38]4 years ago
7 0

Answer:

Step-by-step explanation:

i tried but i cant succed.

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3 years ago
(a-2b)3 when a=-2 and b= -1/2
Aleks [24]

Answer: (a-2b)3=3a-6b

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

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Step-by-step explanation:

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3 years ago
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IrinaK [193]
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3 years ago
What is symmetric property of equality? What is the answer to this problem?
umka21 [38]
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3 years ago
Evaluate the integral. (sec2(t) i t(t2 1)8 j t7 ln(t) k) dt
polet [3.4K]

If you're just integrating a vector-valued function, you just integrate each component:

\displaystyle\int(\sec^2t\,\hat\imath+t(t^2-1)^8\,\hat\jmath+t^7\ln t\,\hat k)\,\mathrm dt

=\displaystyle\left(\int\sec^2t\,\mathrm dt\right)\hat\imath+\left(\int t(t^2-1)^8\,\mathrm dt\right)\hat\jmath+\left(\int t^7\ln t\,\mathrm dt\right)\hat k

The first integral is trivial since (\tan t)'=\sec^2t.

The second can be done by substituting u=t^2-1:

u=t^2-1\implies\mathrm du=2t\,\mathrm dt\implies\displaystyle\frac12\int u^8\,\mathrm du=\frac1{18}(t^2-1)^9+C

The third can be found by integrating by parts:

u=\ln t\implies\mathrm du=\dfrac{\mathrm dt}t

\mathrm dv=t^7\,\mathrm dt\implies v=\dfrac18t^8

\displaystyle\int t^7\ln t\,\mathrm dt=\frac18t^8\ln t-\frac18\int t^7\,\mathrm dt=\frac18t^8\ln t-\frac1{64}t^8+C

8 0
4 years ago
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