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fenix001 [56]
3 years ago
7

Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the corre

ct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers.
For the right triangle shown, the lengths of two sides are given. Find the third side. Leave your answer in simplified, radical form.

a = 5, b = 10, c =

Mathematics
1 answer:
MArishka [77]3 years ago
4 0

Answer:

5 sqrt(5) =c

Step-by-step explanation:

We can use the Pythagorean theorem to find the length of the hypotenuse

a^2 + b^2 = c^2 since this is a right triangle

5^2 + 10^2 = c^2

25+100 = c^2

125 = c^2

Take the square root of each side

sqrt(125) = sqrt(c^2)

sqrt(25*5) = c

sqrt(25) sqrt(5) = c

5 sqrt(5) =c

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1.1(1.6x+0.4)-0.44= -17.6
sertanlavr [38]

Answer:

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

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Polynomial Equation question<br>(Screenshot included)
Illusion [34]

y = {x}^{3}  - 4 {x}^{2}  - 45x \\ y = x( {x}^{2}  - 4x - 45) \\ y = x(x + 5)(x - 9) \\ x = 0, - 5,9
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Evaluate c (y + 7 sin(x)) dx + (z2 + 9 cos(y)) dy + x3 dz where c is the curve r(t) = sin(t), cos(t), sin(2t) , 0 ≤ t ≤ 2π. (hin
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Treat \mathcal C as the boundary of the region \mathcal S, where \mathcal S is the part of the surface z=2xy bounded by \mathcal C. We write

\displaystyle\int_{\mathcal C}(y+7\sin x)\,\mathrm dx+(z^2+9\cos y)\,\mathrm dy+x^3\,\mathrm dz=\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r

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By Stoke's theorem, the line integral is equivalent to the surface integral over \mathcal S of the curl of \mathbf f. We have


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so the line integral is equivalent to

\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\mathrm d\mathbf S
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\mathbf s(u,v)=(u\cos v,u\sin v,2u^2\cos v\sin v)=(u\cos v,u\sin v,u^2\sin2v)

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\mathrm d\mathbf S=\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv=(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv

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\displaystyle\iint_{\mathcal S}(-2u^2\sin2v,-3u^2\cos^2v,-1)\cdot(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv
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