What are the solutions to the quadratic equation 5x2 – 8= –12x?
2 answers:
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Answer:
The figure is a straight line.
The figure lies in the first and the third quadrants.
Step-by-step explanation:
Take five ordered pairs as
Here, the first and second coordinates are equal.
Now, plot these points and connect them.
From the graph, it can be observed that the figure is a straight line.
Also, the figure lies in the first and the third quadrants.
Answer:
(a) -69.59 . . . . the approximate slope of the curve at x=2
(b) 26.583 . . . . the value of x where the slope of f(x) is about -25.
Step-by-step explanation:
(a) The slope of a function <em>f </em>over the interval [x, x+h] is given by ...
... slope = (f(x+h) -f(x))/h
The problem statement asks for the value of ...
... (f(2.01) -f(2))/0.01
Matching this expression to the pattern above, we find x = 2, h = 0.01. An appropriate interpretation of the expression is that it gives the approximate slope of f(x) near x=2.
The value of that slope is about ...
... (f(2.01) -f(2))/0.01 ≈ 1850·(0.92122386 -0.9216)/.01 ≈ -69.5857
(b) f(x+1) = 1850·0.96^(x+1) = 0.96·f(x)
Then the equation becomes ...
... 0.96·f(x) = f(x) -25
... 25 = 0.04·f(x) . . . . . . . . add 25-0.96·f(x)
... 625 = f(x) . . . . . . . . . . . multiply by 25
... 625 = 1850·0.96^x . . . substitute definition of f(x)
... 625/1850 = 0.96^x . . . divide by the coefficient of the exponential term
... log(625/1850) = x·log(0.96) . . . . take logs
... x = log(625/1850)/log(0.96) ≈ -0.471292/-0.0177288
... x ≈ 26.583
If you rearrange the equation to ...
... f(x+1) -f(x) = -25
Comparing this to the equation for slope, above, it looks a lot like we're trying to find the interval [x, x+1] over which the average slope is -25.
