Answer:
A. 
Step-by-step explanation:
So to get the area of a square, we need to find the length of one side.
We know the length of the larger square is a, so the area of the larger cube is 
We can find the length of a side of the smaller square by using pythagoreans theorem to find the hypotenuse of the triangle formed in the bottom left corner. The length of one side along the x axis is a - b, and the length of the other side, along the y-axis, is b.
We can plug it into pythagoreans theorem to get
(C represents the length of one side of the smaller square, and the hypotenuse of the triangle)
The area of the smaller triangle is C squared to the area of the smaller triangle is
To get the ratio of the smaller square in comparison to the larger square we divide the area of the smaller square by the area of the larger square.
So the ratio should be
Recall that
d/dx sech(x) = - sech(x) tanh(x)
d/dx tan⁻¹(x) = 1/(1 + x²)
Then by the chain rule,
dy/dx = - sech(x) tanh(x) / (1 + x²)
Answer: all real numbers
Step-by-step explanation:
the domain for any function is always, all real numbers or in other words, negative infinity to positive infinity.
The normal vector to the plane <em>x</em> + 3<em>y</em> + <em>z</em> = 5 is <em>n</em> = (1, 3, 1). The line we want is parallel to this normal vector.
Scale this normal vector by any real number <em>t</em> to get the equation of the line through the point (1, 3, 1) and the origin, then translate it by the vector (1, 0, 6) to get the equation of the line we want:
(1, 0, 6) + (1, 3, 1)<em>t</em> = (1 + <em>t</em>, 3<em>t</em>, 6 + <em>t</em>)
This is the vector equation; getting the parametric form is just a matter of delineating
<em>x</em>(<em>t</em>) = 1 + <em>t</em>
<em>y</em>(<em>t</em>) = 3<em>t</em>
<em>z</em>(<em>t</em>) = 6 + <em>t</em>
