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3 years ago

Airport A is located on a coordinate plane at (-18, 14). Airport B is located at (8, 14). How far apart are the airports?

Mathematics

1 answer:

Setler [38]3 years ago

4 0

Answer:

26 units

Step-by-step explanation:

Use the distance formula, d = $\sqrt{(x2 - x1)^2 + (y2 - y1)^2$

Plug in the points:

d = $\sqrt{(-18 - 8)^2 + (14 - 14)^2$

Simplify:

d = $\sqrt{(-26^2 + (0)^2}$

d = $\sqrt{676}$

d = 26

So, the airports are 26 units apart

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The sum of two numbers is 64 and the difference is 6. What are the numbers?

Ne4ueva [31]

Answer:

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Step-by-step explanation:

x+y=64

x-y=6

Solve by substitution.

x=6+y

6+y+y=64

6+2y=64

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Then plug in 29 for y.

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2 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:

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

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

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

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3 years ago

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max2010maxim [7]

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Bogdan [553]

Answer is B.
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