Answer:
155.7
Step-by-step explanation:
Use what you know.
Segment AC is 130 ft
Segment CD is 70ft
If you use the Pythagorean Theorem, in this case being = +
To find segment CE, you would do r-70
So, = +
= 16,900 + -140r + 4900
Add the -140r to the left side and then get rid of the two . Then Add 16,900 and 4900 together
You'll end up with 140r = 21,800
Divide 140 on each side.
Your final answer will be 155.7 (rounded to the nearest tenth)
I hope this helps.....
Explanation:
I took this in 9th grade so I know how to do this kind of stuff
![\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ h=-16t^2+\stackrel{\stackrel{v_o}{\downarrow }}{65}t](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Binitial%20velocity%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20~~~~~~%5Ctextit%7Bin%20feet%7D%20%5C%5C%5C%5C%20h%28t%29%20%3D%20-16t%5E2%2Bv_ot%2Bh_o%20%5Cend%7Barray%7D%20%5Cquad%20%5Cbegin%7Bcases%7D%20v_o%3D%5Cstackrel%7B%7D%7B%5Ctextit%7Binitial%20velocity%20of%20the%20object%7D%7D%5C%5C%5C%5C%20h_o%3D%5Cstackrel%7B%7D%7B%5Ctextit%7Binitial%20height%20of%20the%20object%7D%7D%5C%5C%5C%5C%20h%3D%5Cstackrel%7B%7D%7B%5Ctextit%7Bheight%20of%20the%20object%20at%20%22t%22%20seconds%7D%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20h%3D-16t%5E2%2B%5Cstackrel%7B%5Cstackrel%7Bv_o%7D%7B%5Cdownarrow%20%7D%7D%7B65%7Dt)
now, take a look at the picture below, so for 2) and 3) is the vertex of this quadratic equation, 2) is the y-coordinate and 3) the x-coordinate.
12. Make equation and plug in
13. Proportion
14. Finding a constant rate, and then manipulate
12. Where x is miles driven, and y is answer, and z is days
19.95z+0.15x=y
Plug in two values
19.95 x 2 + 0.15 x 250 = y
39.93+37.5 = y
y=77.43
13. 5:2 as 175:x
5/2 = 175/x
x = 70
70+175=245
14. 260:5 = x:1
260/5=x/1
x = 52
52:1 = y:9
52/1 = y/9
y = 468
I hope you understand the process.
Sodium oxide is a simple strongly basic oxide. It is basic because it contains the oxide ion, O2-, which is a very strong base with a high tendency to combine with hydrogen ions. Reaction with water: Sodium oxide reacts exothermically with cold water to produce sodium hydroxide solution
