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aliina [53]
3 years ago
14

Solve the attached question#yohaniJaman​

Mathematics
1 answer:
aksik [14]3 years ago
5 0

HETY is a parallelogram.

HT and EY are diagonals. We know that diagonals divides the parallelogram into two equal parts.

So ar(HET) = ar(HTY)

And, ar(HEY) = ar(EYT) now, in AHET, diagonal EY bisects the line segment HT and also the AHET,

∴ar(AHOE) = ar(AEOT)

Similarly in AETY

ar(ΔΕΟΤ) = ar(ΔΤΟΥ)

And in AHTY,

ar(ATOY) = ar(AHOY)

That means diagonals in parallelogram divides it into four equal parts.

Hence Proofed.

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Find the curl of ~V<br> ~V<br> = sin(x) cos(y) tan(z) i + x^2y^2z^2 j + x^4y^4z^4 k
ch4aika [34]

Given

\vec v =  f(x,y,z)\,\vec\imath+g(x,y,z)\,\vec\jmath+h(x,y,z)\,\vec k \\\\ \vec v = \sin(x)\cos(y)\tan(z)\,\vec\imath + x^2y^2z^2\,\vec\jmath+x^4y^4z^4\,\vec k

the curl of \vec v is

\displaystyle \nabla\times\vec v = \left(\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z}\right)\,\vec\imath - \left(\frac{\partial h}{\partial x}-\frac{\partial f}{\partial z}\right)\,\vec\jmath + \left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\,\vec k

\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ - \left(4x^3y^4z^4-\sin(x)\cos(y)\sec^2(z)\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k

\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ + \left(\sin(x)\cos(y)\sec^2(z)-4x^3y^4z^4\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k

7 0
3 years ago
Total of 49 1/4 + 3 3/8
Ede4ka [16]
49*4=196+1=197/4 
<span>8*3=24+3=27/8 </span>

<span>then make the denominator the same</span>
<span>multiply 197*2 and 4*2 = 394/8 </span>
<span>then 394/8 +27/8 =421/8 </span>
<span>u can either leave it here or in decimal form which would be 52.625 or in mixed numbers-52 5/8</span>
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Answer:

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Step-by-step explanation:

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Kryger [21]

Answer:

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Step-by-step explanation:


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3 years ago
Can you help with this question?
Ilia_Sergeevich [38]
It is 0.525 that is what it is 
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