Hhczfik in nb vguiigdr 5th hb. jigfddtjnbvt
Answer:
Hi. The subexpression is:
1/![x^{5}+x^{6}+x^{3}](https://tex.z-dn.net/?f=x%5E%7B5%7D%2Bx%5E%7B6%7D%2Bx%5E%7B3%7D)
Step-by-step explanation: Adding the exponents of x the equation is as follows
:
![x^{5}+x^{6}+x^{3}](https://tex.z-dn.net/?f=x%5E%7B5%7D%2Bx%5E%7B6%7D%2Bx%5E%7B3%7D)
if you multiply by the subexpression 1/
, the equation would be simplified, as follow:
(
)*(1/
) = 1
Answers:
x = 3y/2
y = 2/3 x
which ever variable you are looking for
hope i helped :) <span />
Answer:
115m
Step-by-step explanation:
Note that one of the face of a pyramid is a right angled triangle.
Since the pyramid was about 147m tall, the height of triangle will be 147m. If one of the faces is built with one of its faces at 52° to the incline, we can find the original length of one of its sides using SOH CAH TOA
tan(theta) = opposite/adjacent
Given theta = 52°, opposite = 147m (the side directly opposite the angle)
Substituting we have;
Tan52° = 147/adjacent
Adjacent = 147/tan52°
Adjacent = 114.8m
= 115m
Therefore 115m is the original length of one of its side
Answer:
![a.\ \ \ A=684\ m^2](https://tex.z-dn.net/?f=a.%5C%20%5C%20%5C%20A%3D684%5C%20m%5E2)
![b.\ \ \ A=130.2 \ km^2](https://tex.z-dn.net/?f=b.%5C%20%5C%20%5C%20A%3D130.2%20%5C%20km%5E2)
Step-by-step explanation:
a. Area is calculated by summing the areas of the prism's individual surfaces.
#First, calculate the areas of the right-angled surfaces:
![Area=2(\frac{1}{2}bh), b=9, h=12\\\\=2\times \frac{1}{2}\times 9\times 12\\\\=108\ m^2](https://tex.z-dn.net/?f=Area%3D2%28%5Cfrac%7B1%7D%7B2%7Dbh%29%2C%20b%3D9%2C%20h%3D12%5C%5C%5C%5C%3D2%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%209%5Ctimes%2012%5C%5C%5C%5C%3D108%5C%20m%5E2)
#We then find the areas of the rectangular surfaces:
![A=lw\\\\=15\times 16+9\times 16+16\times 12\\\\=576\ m^2](https://tex.z-dn.net/?f=A%3Dlw%5C%5C%5C%5C%3D15%5Ctimes%2016%2B9%5Ctimes%2016%2B16%5Ctimes%2012%5C%5C%5C%5C%3D576%5C%20m%5E2)
#We sum the areas to find the total surface areas:
![A=\sum{Areas_i}\\\\=108+576\\\\=684\ m^2](https://tex.z-dn.net/?f=A%3D%5Csum%7BAreas_i%7D%5C%5C%5C%5C%3D108%2B576%5C%5C%5C%5C%3D684%5C%20m%5E2)
Hence, the prism's surface area is ![684\ m^2](https://tex.z-dn.net/?f=684%5C%20m%5E2)
b.Area is calculated by summing the areas of the prism's individual surfaces.
#First, calculate the areas of the right-angled surfaces:
![A=2\times \frac{1}{2}bh,\ b=12, h=4.1\\\\=2\times 0.5\times 12\times 4.1\\\\=49.2 \ km^2](https://tex.z-dn.net/?f=A%3D2%5Ctimes%20%5Cfrac%7B1%7D%7B2%7Dbh%2C%5C%20b%3D12%2C%20h%3D4.1%5C%5C%5C%5C%3D2%5Ctimes%200.5%5Ctimes%2012%5Ctimes%204.1%5C%5C%5C%5C%3D49.2%20%5C%20km%5E2)
#We then find the areas of the rectangular surfaces:
![A=lw\\\\=5\times 3+3\times 10+12\times 3\\\\=81\ km^2](https://tex.z-dn.net/?f=A%3Dlw%5C%5C%5C%5C%3D5%5Ctimes%203%2B3%5Ctimes%2010%2B12%5Ctimes%203%5C%5C%5C%5C%3D81%5C%20km%5E2)
#We sum the areas to find the total surface areas:
![A=A_r+A_t\\\\=81+49.2\\\\=130.2 \ km^2](https://tex.z-dn.net/?f=A%3DA_r%2BA_t%5C%5C%5C%5C%3D81%2B49.2%5C%5C%5C%5C%3D130.2%20%5C%20km%5E2)
Hence, the prism's surface area is ![130.2 \ km^2](https://tex.z-dn.net/?f=130.2%20%5C%20km%5E2)