Which expression is equivalent to (a6b-³) 1/3

Mathematics

1 answer:

tamaranim1 [39]3 years ago

3 0

Answer:

Step-by-step explanation:

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Brainliest + Points<br><br> Help please and explain!

Olin [163]

ADD = 300

ITI = 323

ON - 33 (Almost a 6 at the end!)

4 0

3 years ago

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive an

sleet_krkn [62]

Answer:

-6re−r [sin(6θ) - cos(6θ)]

Step-by-step explanation:

the Jacobian is ∂(x, y) /∂(r, θ) = δx/δθ × δy/δr - δx/δr × δy/δθ

x = e−r sin(6θ), y = er cos(6θ)

δx/δθ = -6rcos(6θ)e−r sin(6θ), δx/δr = -sin(6θ)e−r sin(6θ)

δy/δθ = -6rsin(6θ)er cos(6θ), δy/δr = cos(6θ)er cos(6θ)

∂(x, y) /∂(r, θ) = δx/δθ × δy/δr - δx/δr × δy/δθ

= -6rcos(6θ)e−r sin(6θ) × cos(6θ)er cos(6θ) - [-sin(6θ)e−r sin(6θ) × -6rsin(6θ)er cos(6θ)]

= -6rcos²(6θ)e−r (sin(6θ) - cos(6θ)) - 6rsin²(6θ)e−r (sin(6θ) - cos(6θ))

= -6re−r (sin(6θ) - cos(6θ)) [cos²(6θ) + sin²(6θ)]

= -6re−r [sin(6θ) - cos(6θ)] since [cos²(6θ) + sin²(6θ)] = 1

6 0

3 years ago

A particular circle in the standard (x,y) coordinate

matrenka [14]

For any circle with Cartesian equation
$(x-a)^2 + (y-b)^2 = r^2$ ,
we have that the centre of the circle is $(a,b)$ , and the radius of the circle is $r$ .

So in the case that
$(x-5)^2 + y^2 = 38$ ,
we essentially have that
$a = 5, b = 0, r^2 = 38$ .

So the centre of the circle is $(5,0)$ , and the radius is $\sqrt{38}$ .