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2 years ago

8 with a exponent of 2 +7

Mathematics

2 answers:

Lyrx [107]2 years ago

8 0

Answer:

Step-by-step explanation:

8²=64

64+7=71

I hope this helps!

goblinko [34]2 years ago

8 0

Answer: soooo 8^2+(7)? if so the answer is 71

Step-by-step explanation: 8 times 8 is 64. add 7 to 64 and get 71

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2 years ago

Read 2 more answers

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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3 years ago

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Veseljchak [2.6K]

Answer:

Step-by-step explanation:

f(x) (8x² - 3x + 1) / (2x² + 2x +3)

the first thing I did was to graph the function to see if there was a horizonal asymptote using Desmos graphing calculator and there was and it was 4

you want to take the limit as x approaches infinity

essentially lim(x to ∞) of f(x) = 8x²/ 2x² is the dominate terms

lim(x to ∞) of f(x) = 8 / 2

lim(x to ∞) of f(x) = 4

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3 years ago