Answer:
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The main entrance to the Louvre art museum is shaped like a pyramid. The
1 answer:
The Lateral Surface Area of Square based pyramid whose height= 71 feet Slant height= 90 feet and Base= 115 feet is 20700 .
The Lateral Surface Area is 20700
The word "lateral" means "belonging to the side". So the lateral area of a square pyramid is the sum of the areas of its side faces. This is also known as the lateral surface area (LSA) of the square pyramid. We know that a square pyramid has:
- a base which is a square.
- 4 side faces, each of which is a triangle.
It is given that,
- height, h= 71 feet
- Slant height, l= 90 feet
- Base, b= 115 feet
Lateral Surface Area of square pyramid is,
LSA=
Where p= perimeter of base
l= slant height
Perimeter of base = 4 x side
= 4 x 115
= 460 feet
Now, Lateral Surface Area of square pyramid is,
=
= 460 x 45
= 20700
Hence the Lateral Surface Area of square pyramid is 20700
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<h3>Answer: The average rate of change for both is -2</h3>
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Explanation:
The x interval [0,3] is the same as writing
It starts at x = 0 and ends at x = 3.
The graph shows that x = 0 leads to y = 3. So we have the point (0,3) on the parabola. We also have the point (3,-3) on the parabola.
Let's find the slope of the line through these endpoints.
The slope is -2. This is the average rate of change from x = 0 to x = 3.
This is because:
slope = rise/run = (change in y)/(change in x) = average rate of change.
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Now let's find the slope for the table.
Focus on the rows for x = 0 and x = 3. They lead to f(x) = 10 and f(x) = 4 respectively.
We have (0,10) and (3,4) as our two points this time.
We get the same slope as before, so we have the same rate of change.
Notice the change in y (-6) is the same as before. So we could pick any two y values we want as long as there's a gap of 6 between them, and the second y value is smaller than the first.
To find the gradient of the tangent, we must first differentiate the function.
The gradient at x = 0 is given by evaluating f'(0).
The derivative of the function at this point is negative, which tells us <em>the function is decreasing at that point</em>.
The tangent to the line is a straight line, so we will have a linear equation of the form y = mx + c. We know the gradient, m, is equal to -1, so
Now we need to substitute a point on the tangent into this equation to find c. We know a point when x = 0 lies on here. To find the y-coordinate of this point we need to evaluate f(0).
So the point (0, -1) lies on the tangent. Substituting into the tangent equation: