All answers are the pic.
When you are lining up the ratios, make sure they are aligning each other with the same unit.
When you are lining up the ratios, make sure they are aligning each other with the same unit.
Tyler needs at least $205 for a new video game system. He has already saved $30. He earns $7 an hour at hisjob. Write and solve
1 answer:
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To determine the maximum value of a quadratic function opening downwards, we are going to find the vertex; then the y-value of the vertex will be our maximum.
To find the vertex (h,k) (where h=x-coordinate and k=y-coordinate) of a quadratic function of the form
we'll use the vertex formula: 
, and then we are going to replace that value in our original function to find k.
So, in our function
, 
and 
.
Lets replace those values in our vertex formula:
Now that we know the x-coordinate of our vertex, lets replace it in the original function, to get the y-coordinate:
We just prove that the vertex of
is (2,1), and for the graph we can tell that the vertex of 
is (-2,4). The only thing left is compare their y-coordinates to determine w<span>hich one has the greater maximum value. Since 4>1, we can conclude that </span> has the greater maximum.
To find the vertex (h,k) (where h=x-coordinate and k=y-coordinate) of a quadratic function of the form
So, in our function
Lets replace those values in our vertex formula:
Now that we know the x-coordinate of our vertex, lets replace it in the original function, to get the y-coordinate:
We just prove that the vertex of
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Answer:
see attachment
Step-by-step explanation:
The iterator for Newton's method gives the next approximation (x') as ...
x' = x -g(x)/g'(x) . . . . . where g'(x) is the derivative of g(x)
We have defined g(x) = x^3+x+3, the function we want the zero of. We have defined the iteration function to be f(x).
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<em>Additional comment</em>
Modern graphing calculators not only make the iteration trivially simple, they also give a first approximation good to 2 or 3 decimal places in many cases.
