Answer:
-270 to -360 degrees I quadrant
And for this case since -300° is between -270 to -360 degrees we can conclude that this angle is in the I quadrant
Step-by-step explanation:
For this case we can solve this problem taking in count the following rule when we have negative angles:
0 to -90 degrees IV quadrant
-90 to -180 degrees III quadrant
-180 to -270 degrees II quadrant
-270 to -360 degrees I quadrant
And for this case since -300° is between -270 to -360 degrees we can conclude that this angle is in the I quadrant
Lateral Area, namely the area of its sides, namely excluding its base.
well, the pyramid is standing on one of its triangular faces, so we'll have to exclude that.
now, let's notice, is a square pyramid, so it has an 8x8 square, and it has triangular faces that have a <u>base of 8 and a height of 22</u>.
![\bf \stackrel{\textit{square's area}}{(8\cdot 8)}~~+~~\stackrel{\textit{3 triangular faces' area}}{3\left[ \cfrac{1}{2}(8)(22) \right]}\implies 64+264\implies 328](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Bsquare%27s%20area%7D%7D%7B%288%5Ccdot%208%29%7D~~%2B~~%5Cstackrel%7B%5Ctextit%7B3%20triangular%20faces%27%20area%7D%7D%7B3%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%288%29%2822%29%20%5Cright%5D%7D%5Cimplies%2064%2B264%5Cimplies%20328)
127+172=299 so Ryan read 299 pages in total from monday to tuesday
Answer:
your formula for 1 is L x W x H which would be 6 x 4 x 5
your formula for 2 is L x W x H otherwise known as 2 x 1.5 x 3
your formula for 3 is L x W x H which would be 10 x 10 x 30
your formula for 4 is L x W x H or 6 x 3 x 4
I don't know what it means by plug in.
