The correct answer is: "Graph 1" ; that is, the "first graph to the left" .
________________________________________________________
→ {among the two graphs provided as answer choices.}.
________________________________________________________
<u>Note</u>:
________________________________________________________
The line drawn on the graph shown on the "first graph" — (that is, the graph shown to the far left, among the two graphs provided as answer choices) — reflect the linear equation given:
" y = 6x " ;
and the line shown on the "first graph shown to the left" consists of the coordinates reflected on the table given:
" (0, 0) , (1, 6), (2, 12), (3, 18) " .
___________________________________________________
Note that the "second graph, to the far right, is incorrect; the line provided in the "second graph" shows that:
when "x = 4" , "y = 20" .
This is incorrect; since:
Given the equation:
" y = 6x " ;
→ y = 6(4) ;
→ y = 24 ; NOT "20" ; so "Graph 2" is incorrect.
___________________________________________________
The correct answer is: "Graph 1" ; that is, the "first graph to the left" .
___________________________________________________
Remark.
The problem is a bit indistinct. Where exactly are the two edges of the road? I'm going to say that they are the x intercepts, but that may not be true. Certainly it does not have to be true at all.
Graph.
A graph has been made for you. The maximum is marked for you. It is an approximation The actual height can be more accurately found.
Height
y = (-1/200)(x - 16)(x + 16)
y = (-1/200)*(x^2 - 256)
The maximum height for this graph only is when x = 0.Other graphs require completing the square.
y = (-1/200) * (-256)
y = 1.28 exactly. I thought the graph might be rounding the answer. It is not.
It’s 3/12 you just need to find common multiples
Answer:
The distance cover by Bell is 8 miles .
Step-by-step explanation:
Given as :
The distance cover by Mike = The distance cover by Pete = D miles
Let The distance cover by Bell = d miles
The distance cover by Mike = 15 miles + distance cover by Bell
i.e D = 15 + d ............1
Again
The distance cover by Pete = three time distance cover by Bell - 1 miles
i.e D = 3 × B - 1
Or, D = 3 d - 1 ............2
Now, Solving equation 1 and 2
So, 15 + d = 3 d - 1
Or, 3 d - d = 15 + 1
Or, 2 d = 16
∴ d = 
i.e d = 8 miles
So,The distance cover by Bell = d = 8 miles
Hence, The distance cover by Bell is 8 miles . Answer
Answer:
x=5
Step-by-step explanation: