Answer:
D. 15°
Step-by-step explanation:
EGF and HGF are equal and 75°
since JGH is 90° we can say JGI is 90 - 75 = 15°
The key features of a quadratic graph that can identified are; x and y intercepts, axis of symmetry and vertex
<h3>Keys features of a quadratic graph</h3>
The key features are the x-intercepts, y-intercepts, axis of symmetry, and the vertex.
If we add units we can move this function upwards, downwards leftwards and rightwards.
- If we add a positive number to the x-variable, then the graph will move to the left.
- If we add a negative number to the x-variable, then the graph will move to the right.
- If we add a positive number to y-variable, then the graph will move upwards.
- If we add a negative number to y-variable, then the graph will move downwards.
Hence, if we compare the rules we use before with linear function, there's no distinction between horizontal and vertical movements, because if we add to x-variable, then y-variable will be also affected.
Learn more about quadratic graphs here:
brainly.com/question/1214333
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the assumption being that the endpoints are two continuous points in the pentagon, Check picture below.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-1}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{[2-(-1)]^2+[3-4]^2}\implies d=\sqrt{(2+1)^2+(3-4)^2} \\\\\\ d=\sqrt{9+1}\implies d=\sqrt{10}~\hfill \stackrel{\stackrel{~\hfill \stackrel{\textit{5 sides}}{}}{\textit{perimeter of the pentagon}}}{5\sqrt{10}}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B-1%7D~%2C~%5Cstackrel%7By_1%7D%7B4%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B2%7D~%2C~%5Cstackrel%7By_2%7D%7B3%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%3D%5Csqrt%7B%5B2-%28-1%29%5D%5E2%2B%5B3-4%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%282%2B1%29%5E2%2B%283-4%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%3D%5Csqrt%7B9%2B1%7D%5Cimplies%20d%3D%5Csqrt%7B10%7D~%5Chfill%20%5Cstackrel%7B%5Cstackrel%7B~%5Chfill%20%5Cstackrel%7B%5Ctextit%7B5%20sides%7D%7D%7B%7D%7D%7B%5Ctextit%7Bperimeter%20of%20the%20pentagon%7D%7D%7D%7B5%5Csqrt%7B10%7D%7D)
SA=2(lw+wh+lh) This is the formula for finding the surface area of a rectangular prism, where SA is surface area, l is length, w is width, and h is height.
208=2(lw+wh+lh)
104=lw+wh+lh Here, I divided both sides by 2 to get ride of the 2.
Now, I used prime factorization to find out all the prime factors of 104, which are 2, 2, 2, and 13. Since rectangular prisms only have 3 dimensions, I needed to combine two of the prime factors. In this case, I can either combine 2 of the 2s to get 2, 4, and 13 or I can combine 13 with one of the 2s to get 26, 2, and 2.
If my dimensions were 2, 4, and 13...
my surface area would be 172 sq cm.
If my dimensions were 2, 2, and 26...
my surface area would be 208 sq cm.
Hence, the width of the rectangular prism when the surface area is 208 square centimeters can be either 2 or 26.
