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3 years ago

HELP PLEASEWhat is the value of the exponential expression below?

Mathematics

2 answers:

vladimir2022 [97]3 years ago

6 0

Hi, I got C because you just put 1/2 in for an exponent.. so, 16 of an exponent of 1/2.. it would be letter C. 4

babunello [35]3 years ago

4 0

Square root of 16. Equals 4

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strojnjashka [21]

Answer:

Cody was 148.5cm on the first day of school last year.

Step-by-step explanation:

10% = 0.1

165 x 0.1 = 16.5

165 - 16.5 = 148.5

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3 years ago

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Kamila [148]

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3 years ago

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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

Factor the following polynomial. ( Correct and best explained answer gets brainliest. )

mamaluj [8]

Glad you asked this question! Over the years of working with my professor we learned that if phs= 5 (constant) times the j to the power of 2 you will get the formula or polynomialic equation. It quite fascinating and due to using his equation the answer in fact is "c". Cheers

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3 years ago