(do it if you want the original height of khafre's pyramid, next to the great pyramid, was about 471 ft. each side of its square base was about 708 ft. to the nearest square foot, what is the lateral area of a pyramid with those dimensions?  

We have the base measurement for the triangles composing the sides. We need their altitude.  

Imagine a right triangle with these sides:  

(1) from the apex of the pyramid to the point on the base directly underneath it (471 ft);  

(2) from the middle of base edge to the point underneath the apex (half of 708 ft = 354 ft);  

(3) the hypotenuse, the altitude of a triangular side.  

Using the Pythagorean Theorem, we find the hypotenuse to be  

√(471^2 + 354^2) = about 589.2 feet  

Now we add up the four triangles' areas. Each is base * altitude / 2, so:  

4 x 708 x 589.2 / 2  

= 2 x 708 x 589.2  

= 834,307.2 square feet, the lateral area.

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

mark brainliest!