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3 years ago

What is the absolute value of |-64]

Mathematics

1 answer:

Otrada [13]3 years ago

3 0

Answer:

Step-by-step explanation:

Step 1:

Absolute Value is the distance between a number and 0.

Answer:

The answer is 64 because - 64 is still 64 units away from zero.

Hope This Helps :)

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Step-by-step explanation:

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3 years ago

Write the coordinates of the vertices after a reflection over the x-axis. 1072

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The reflection about x-axis results in same x coodinate with oppositive of y coordinate. It can be expressed as,

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$Q(6,-8)\rightarrow Q^{\prime}(6,8)$ $R(7,-8)\rightarrow R^{\prime}(7,8)$ $S(7,-5)\rightarrow Q^{\prime}(7,5)$ $T(6,-5)\rightarrow T^{\prime}(6,5)$

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3 years ago

Read 2 more answers

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

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

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

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

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3 years ago