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2 years ago

Please help me with this problem I need your help badly

Mathematics

1 answer:

Lelechka [254]2 years ago

7 0

8.223 inches.

38.000
-29.777
8.223

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Calculate the area of the square shown on the xy-plane in the diagram.

natka813 [3]

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Step-by-step explanation:

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2 years ago

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The first term in a sequence is 84. The pattern follows the rule add 24. What are the next four terms?

weqwewe [10]

Answer: 108, 132, 156, and 180

Step-by-step explanation:

If you need to add 24 to the first term for the next 4 terms, you would have

84 + 24 = 108

108 + 24 = 132

132 + 24 = 156

156 + 24 = 180

So your sequence would be 108, 132, 156, and 180 for the next 4 terms.

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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:

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

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

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

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2 years ago