Given the relation x=y4-9y2, identify the x- and y- axis intercepts
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So...the tent, seems to be the first picture below
the second picture, is just a break-down of the pyramid
so... that's the value of "a" and the "c" or "slant height", namely the missing slanted side on that second picture
we need the slant height, in order to get the area of that triangular face, and the base of that triangle is, just "a"
since the area of a triangle is 1/2 bh, and the base is "a" and the altitude or height here is the slant height "c", then the area of that triangular face is
and the area of the lateral sides for the prism, are " a * 1.5" or
so, you have 4 triangles, and 4 rectangles to be added, so.. .let's do that, to get the canvas then
![\bf \begin{array}{clclll} 4\left[ \left( \cfrac{1}{2} \right)\left( \sqrt{4.9}\right)\left( \sqrt{\cfrac{89}{40}} \right) \right]&+&4[(\sqrt{4.9})(1.5)]\\ \uparrow &&\uparrow \\ \textit{4 triangles}&&\textit{4 rectangles} \end{array}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bclclll%7D%0A4%5Cleft%5B%20%5Cleft%28%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5Cleft%28%20%5Csqrt%7B4.9%7D%5Cright%29%5Cleft%28%20%5Csqrt%7B%5Ccfrac%7B89%7D%7B40%7D%7D%20%5Cright%29%20%5Cright%5D%26%2B%264%5B%28%5Csqrt%7B4.9%7D%29%281.5%29%5D%5C%5C%0A%5Cuparrow%20%26%26%5Cuparrow%20%5C%5C%0A%5Ctextit%7B4%20triangles%7D%26%26%5Ctextit%7B4%20rectangles%7D%0A%5Cend%7Barray%7D)
that'd be the area of the canvas, or namely, the surface area
notice, we are not including the base of the pyramid, because that's inside the tent, and excluding the base of the prism, because that'd be the ground, and the tent may not include that, well, depends on the tent, in this case, I think a tent this big may not, other smaller ones do though
the second picture, is just a break-down of the pyramid
so... that's the value of "a" and the "c" or "slant height", namely the missing slanted side on that second picture
we need the slant height, in order to get the area of that triangular face, and the base of that triangle is, just "a"
since the area of a triangle is 1/2 bh, and the base is "a" and the altitude or height here is the slant height "c", then the area of that triangular face is
and the area of the lateral sides for the prism, are " a * 1.5" or
so, you have 4 triangles, and 4 rectangles to be added, so.. .let's do that, to get the canvas then
that'd be the area of the canvas, or namely, the surface area
notice, we are not including the base of the pyramid, because that's inside the tent, and excluding the base of the prism, because that'd be the ground, and the tent may not include that, well, depends on the tent, in this case, I think a tent this big may not, other smaller ones do though
