What’s the answerrrr???

Mathematics

1 answer:

Zielflug [23.3K]3 years ago

8 0

J L K BECAUSE JLK IS JLK AND IT WORKS OUT GREAT

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If f is a scalar field and F, G are vector fields, then f F, F · G, and F × G are defined by the following. (f F)(x, y, z) = f(x

valentina_108 [34]

Answer:

Step-by-step explanation:

Consider curl $(fF)$ where $f$ is a scalar function and F is a vector function

$F=F_{1}i +F_{2}j +F_{3}k$

i j k

$curl(fF) = \frac{\partial}{\partial x}$ $\frac{\partial}{\partial y}$ $\frac{\partial}{\partial z}$

$fF_{1}$ $fF_{2}$ $fF_{3}$

$curl(fF)=i(\frac{\partial}{\partial y}(fF_{3})-\frac{\partial}{\partial z}(fF_{2}))-j(\frac{\partial}{\partial x}(fF_{3})-\frac{\partial}{\partial z}(fF_{1}))+k(\frac{\partial}{\partial x}(fF_{2})-\frac{\partial}{\partial y}(fF_{1}))$

$=i(\frac{\partial}{\partial y}(F_{3})+\frac{\partial}{\partial y}(f)-\frac{\partial}{\partial z}(F_{2})-\frac{\partial}{\partial z}(f))-j(\frac{\partial}{\partial x}(F_{3})+\frac{\partial}{\partial x}(f)-\frac{\partial}{\partial z}(F_{1})-\frac{\partial}{\partial z}(f))+k(\frac{\partial}{\partial x}(F_{2})+\frac{\partial}{\partial x}(f)-\frac{\partial}{\partial y}(F_{1})-\frac{\partial}{\partial y}(f))$

$curl(fF)=f(\Delta\times F)+(\Delta f)\times F$

3 0

2 years ago

Help due soon plss!!!!

Gelneren [198K]

Answer:

Cube=150cm^2

Triangular Prism=228ft^2

Step-by-step explanation:

Cube

$A=6a^2=6*5^2=150cm^2$

Triangular prism

Using these formulas

$A=2A_{B}+(a+b+c)h\\A_{B} =\sqrt{s(s*a)(s*b)(s*c)} \\s=\frac{a+b+c}{2}$

Solving for A

$A=ah+bh+ch+\frac{1}{2} \sqrt{-a^4+2(ab)^2+2(ac)^2-b^4+2(bc)^2-c^4} \\=6.5*11+5*11+6.5*11+\frac{1}{2} *\sqrt{-6.5^4+2*(6.5*5)^2+2*(6.5*6.5)^2-5^4+2*(5*6.5)^2-6.5^4} \\=228ft^2$