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| 08 Aug 2017 12:00 AM |
i aint looking for explanations i just need a smart math dude to do it for me i 100% do not have the time to learn this i just need to get this in so i can pass the class i ain't using this math again lmao im not taking another math until i finish senior year my dude
my steam ID is chawliecharles pls hmu |
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Forenz
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| Joined: 31 Oct 2009 |
| Total Posts: 7481 |
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| 08 Aug 2017 12:07 AM |
| 3d figures help a brudda out |
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| 08 Aug 2017 12:19 AM |
I had a long explanation of the other answers typed out before the other thread got deleted. RIP
I encourage you to read through these arguments so that you can understand how the math works.
General Definitions: LATERAL AREA: The area of all sides excluding the bases. SURFACE AREA: The area of all sides including the bases.
Q2. We have a 4x4x4 cube. A) LATERAL AREA I. A cube has 4 sides that aren't bases. II. The area of each side of the cube is 4*4=16 units squared. III. The sum of 4 sides with an area of 16 is 4*16=64 units squared. IV. The LATERAL AREA of the cube is 64 units squared.
B) SURFACE AREA I. The SURFACE AREA is the sum of all the sides. II. The LATERAL AREA is 64 units squared. III. There are two sides forming the bases that are excluded from the LATERAL AREA. IV. These sides have an area 4*4=16 units squared each. V. There are two sides. 16*2#######t#####a########.####32=96 units squared. VII. The SURFACE AREA is 96 units squared.
Q3. LATERAL AREA of a Rectangular Prism I. We need the last side of the triangle forming the base of the prism. II. This side can be found by solving the following equation: 1. 4^2+b^2=6^2 2. 16+b^2=36 3. b^2=20 4. b=sqrt(20) III. The triangle has sides of 4, 6, and sqrt(20). The prism has a height of 8. IV. The area of all the sides is as follows: a. 4*8=32 units squared b. 6*8=48 units squared c. sqrt(20)*8=8sqrt(20) units squared V. LATERAL AREA is the sum of a, b, and c. 32+48+8sqrt(20)=80+8sqrt(20). VI.#######TERAL AREA is 80+8sqrt(20).
Q4. SURFACE AREA of a Rectangular Prism (from Q3). I. SURFACE AREA is LATERAL AREA + AREA OF BASES II. We have the LATERAL AREA (Q3). III. The AREA of a triangular base is hb/2. IV. We have two bases. hb##########V. The HEIGHT of our triangle is sqrt(20) and the BASE of our triangle is 4.##V####b=4sqrt(20) VII. LATERAL AREA + AREA OF BASES = 80+8sqrt(20) + 4sqrt(20) VIII. SURFACE AREA = 80+12sqrt(20).
The math should be right here.
Again, I highly encourage you to review the math and make sure you understand how I reached my conclusions here. Good luck to you.
-False of UAF |
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| 08 Aug 2017 12:26 AM |
| Oh, it says round to the nearest tenth. Just plug the LATERAL AREA from Q3 and the SURFACE AREA from Q4 into a calculator and round to the tenths place. |
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zman250
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| Joined: 20 Jan 2011 |
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| 08 Aug 2017 12:39 AM |
i thought this was for a girl you know
:thinking: |
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