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3 years ago

Plz i want the answer

Mathematics

1 answer:

DanielleElmas [232]3 years ago

3 0

Answer:

it's answer is E. S = (t² - 2p) / a

Step-by-step explanation:

t² = 2p + as

Making s subject then

as = t² - 2p

s =( t² - 2p) / a

Hope it will help :)

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vlabodo [156]

Answer:

25a-20

Step-by-step explanation:

15a+10a-20=

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3 years ago

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vladimir2022 [97]

Answer: =x4+3x3+2x2−4

Step-by-step explanation:

(x2+3x+2)(x2)−4

Distribute:

=(x2)(x2)+(3x)(x2)+(2)(x2)+−4

=x4+3x3+2x2+−4

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Step-by-step explanation:

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Answer:

Step-by-step explanation:

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2 years ago

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