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Sliva [168]
2 years ago
12

Explain why it might be important to be able to work with and manipulate fractions when working in construction.

Mathematics
1 answer:
nirvana33 [79]2 years ago
4 0

Answer:

Addition and Subtraction with different denominators. If the denominators are different then a common denominator needs to be found.

2 pages.

hope this helps

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?????? I don’t get it
QveST [7]

Answer:

His gain was greater on the first play because if you convert 2 yards to feet you get 6ft and 6 feet is greater than 5 feet which means the football player had a greater gain on the first play.

Step-by-step explanation:

1 yard is 3 ft

2yards x 3=6 ft

6ft> 5ft

first play> Second play

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Deena has 2 pairs of white socks, 3 pairs of black socks, 1 pair of red socks, and 2 pairs of navy socks in her sock drawer. Eac
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25 is the correct answer because the amount of white socks out all of the socks is 2/8 which is equal to 1/4. 1/4 is 25%.

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Step-by-step explanation:

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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

8 0
3 years ago
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