Shaded area = area of the hexagon – area of the pentagon + area of the square – area of the equilateral triangle. This can be obtained by finding each shaded area and then adding them.
<h3>Find the expression for the area of the shaded regions:</h3>
From the question we can say that the Hexagon has three shapes inside it,
Also it is given that,
An equilateral triangle is shown inside a square inside a regular pentagon inside a regular hexagon.
From this we know that equilateral triangle is the smallest, then square, then regular pentagon and then a regular hexagon.
A pentagon is shown inside a regular hexagon.
- Area of first shaded region = Area of the hexagon - Area of pentagon
An equilateral triangle is shown inside a square.
- Area of second shaded region = Area of the square - Area of equilateral triangle
The expression for total shaded region would be written as,
Shaded area = Area of first shaded region + Area of second shaded region
Hence,
⇒ Shaded area = area of the hexagon – area of the pentagon + area of the square – area of the equilateral triangle.
Learn more about area of a shape here:
brainly.com/question/16501078
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False if y=f(x) then x= inverse of f(x) or f^-1(x)
I will show you the steps on how you get that answer and if you have any questions after that let me know and I'd be more than happy to help answer them for you.
The first step for solving (1 + y)² is to use the equation (a + b)² = a² + 2ab + b² to expand the expression.
1² + 2 × 1y + y²
1 raised to any power equals 1,, so remove the power.
1 + 2 × 1y + y²
Calculate the product of 2 × 1y.
1 + 2y + y²
Finally,, use the commutative property to reorder the terms.
y² + 2y + 1
Let me know if you have any further questions.
:)
If
represent a family of surfaces for different values of the constant
. The gradient of the function
defined as
is a vector normal to the surface
.
Given <span>the paraboloid
.
We can rewrite it as a scalar value function f as follows:
The normal to the </span><span>paraboloid at any point is given by:
Also, the normal to the given plane
is given by:
Equating the two normal vectors, we have:
</span>
Since, -1 = 2 is not possible, therefore
there exist no such point <span>
on the paraboloid such that the tangent plane is parallel to the plane 3x + 2y + 7z = 2</span>
.
