Quick brainliest asap!! sin^-1 (1.1976) SHOW YOUR WORK so I can see if it makes sense!!

Mathematics

1 answer:

dsp733 years ago

7 0

Answer:

no solutions

Step-by-step explanation:

We want to find $\theta$ such that $\theta=\sin^{-1}(1.1976)$ .

We can take the sine of both sides of the equation to get $\sin(theta)=\sin(\sin^{-1}(1.1976))=1.1976$ .

We know that the range of the sine function goes from -1 to 1 inclusive, so there is no value $\theta$ such that $\sin{\theta}>1$ .

Therefore, the answer is $\boxed{no~solutions}$ and we're done!

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Find equations of the tangent plane and the normal line to the given surface at the specified point. x + y + z = 8exyz, (0, 0, 8

Dima020 [189]

Let $f(x,y,z)=x+y+z-8e^{xyz}$ . The tangent plane to the surface at (0, 0, 8) is

$\nabla f(0,0,8)\cdot(x,y,z-8)=0$

The gradient is

$\nabla f(x,y,z)=\left(1-8yze^{xyz},1-8xze^{xyz},1-8xye^{xyz}\right)$

so the tangent plane's equation is

$(1,1,1)\cdot(x,y,z-8)=0\implies x+y+(z-8)=0\implies x+y+z=8$

The normal vector to the plane at (0, 0, 8) is the same as the gradient of the surface at this point, (1, 1, 1). We can get all points along the line containing this vector by scaling the vector by $t$ , then ensure it passes through (0, 0, 8) by translating the line so that it does. Then the line has parametric equation

$(1,1,1)t+(0,0,8)=(t,t,t+8)$

or $x(t)=t$ , $y(t)=t$ , and $z(t)=t+8$ .

(See the attached plot; the given surface is orange, (0, 0, 8) is the black point, the tangent plane is blue, and the red line is the normal at this point)