Answer:
D.
Step-by-step explanation:
Recall the slope is change of y over change of x which in this case the change in x is 1 while the change in y is 2 since the 
's are going up by 2 while the 
's are going up by 1.
So the slope is 2.
The one equation with slope 2 is D.
The answer is most likely D.
Let's check.
while 
.
while 
.
while 
.
while 
.
Also notice the is doubled (multiplied by 2) to get .
Complete question :
Suppose the average yearly salary of an individual whose final degree is a master's is $41 thousand less than twice that of an individual whose final degree is a bachelor's. Combined, two people with each of these educational attainments earn $118 thousand. Find the average yearly salary of an individual with each of these final degrees.
Answer:
Bachelors = $53000
Masters = $65000
Step-by-step explanation:
Given that :
Let Salary of bachelor's degree holder = b
Salary of Master's degree holder = 2b - 41000
Master's + bachelor's salary = $118,000
Hence,
2b - 41000 + b = 118,000
3b = 118000 + 41000
3b = 159000
b = 159000 / 3
b = $53,000
Bachelor's degree salary = $53,000
Master's degree salary ; 2b - 41000
2(53000) - 41000 = 65000
The walker's speed is
(913 miles) / (2.5 hours) = <em>365.2 miles per hour</em>
(Hey, I just solve um. I don't get into the philosophy.)
well, first off let's check those two points, we know it's centerd at (-26 , 120) and we also know it passes through (0 , 0), so the distance between those two points is its radius
![~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{0}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{-26}~,~\stackrel{y_2}{120})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{radius}{r}=\sqrt{(~~-26 - 0~~)^2 + (~~120 - 0~~)^2} \implies r=\sqrt{(-26)^2 + (120 )^2} \\\\\\ r=\sqrt{( -26 )^2 + ( 120 )^2} \implies r=\sqrt{ 676 + 14400 } \implies r=\sqrt{ 15076 } \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B0%7D~%2C~%5Cstackrel%7By_1%7D%7B0%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B-26%7D~%2C~%5Cstackrel%7By_2%7D%7B120%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%20%5Cstackrel%7Bradius%7D%7Br%7D%3D%5Csqrt%7B%28~~-26%20-%200~~%29%5E2%20%2B%20%28~~120%20-%200~~%29%5E2%7D%20%5Cimplies%20r%3D%5Csqrt%7B%28-26%29%5E2%20%2B%20%28120%20%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20r%3D%5Csqrt%7B%28%20-26%20%29%5E2%20%2B%20%28%20120%20%29%5E2%7D%20%5Cimplies%20r%3D%5Csqrt%7B%20676%20%2B%2014400%20%7D%20%5Cimplies%20r%3D%5Csqrt%7B%2015076%20%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \hspace{5em}\stackrel{center}{(\underset{-26}{h}~~,~~\underset{120}{k})}\qquad \stackrel{radius}{\underset{\sqrt{15076}}{r}} \\\\[-0.35em] ~\dotfill\\\\ ( ~~ x - (-26) ~~ )^2 ~~ + ~~ ( ~~ y-120 ~~ )^2~~ = ~~(\sqrt{15076})^2 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill (x+26)^2+(y-120)^2 = 15076~\hfill](https://tex.z-dn.net/?f=%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%28x-%20h%29%5E2%2B%28y-%20k%29%5E2%3D%20r%5E2%20%5Chspace%7B5em%7D%5Cstackrel%7Bcenter%7D%7B%28%5Cunderset%7B-26%7D%7Bh%7D~~%2C~~%5Cunderset%7B120%7D%7Bk%7D%29%7D%5Cqquad%20%5Cstackrel%7Bradius%7D%7B%5Cunderset%7B%5Csqrt%7B15076%7D%7D%7Br%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%28%20~~%20x%20-%20%28-26%29%20~~%20%29%5E2%20~~%20%2B%20~~%20%28%20~~%20y-120%20~~%20%29%5E2~~%20%3D%20~~%28%5Csqrt%7B15076%7D%29%5E2%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20%28x%2B26%29%5E2%2B%28y-120%29%5E2%20%3D%2015076~%5Chfill)
Answer:
Step-by-step explanation:
We want to find the square root of 50;
We need to first rewrite 50 as a prime factorization.
We now split the square root to get:
Take square root to get:
To the nearest tenth we have
