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stepladder [879]
3 years ago
10

1/6, 2/5, 3/5, 3/7 from least to greatest.

Mathematics
1 answer:
7nadin3 [17]3 years ago
3 0

Answer:

Order from Least to Greatest

1/6  <  2/5  <  3/7  <  3/5



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Find the area of the figure.
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So the right answer is 8 1/8 ft^2.

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solve for b 1/4(b-8)=2. I hope you have time to answer this. Thank you so much for your time you guys make me so happy
Mademuasel [1]

Answer:

16

Step-by-step explanation:

1/4(b-8)=2

b-8=2/(1/4)

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b-8=8

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b=16

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46.8 divided by 1.2 equals
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Answer: 39

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2 years ago
Why is every prime number greater than 2 an odd number?
Arlecino [84]
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3 years ago
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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:

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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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