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dem82 [27]
3 years ago
10

Eduardo says that an increase from 10 to 40 is 300% therefore a decrease from 40 to 10 must be a 300% decrease do you agree with

eduardo explain your reasoning
Mathematics
1 answer:
attashe74 [19]3 years ago
4 0

Answer:

Eduardo's second statement is wrong.

Step-by-step explanation:

Eduardo says that an increase from 10 to 40 is 300% which is true. Because there is an increase by (40 - 10) = 30 and the percent of increase is calculated with respect to 10.  

So, % of increase = \frac{30 \times 100}{10} = 300%.

But when we consider a decrease from 40 to 10, then the decrease is by (40 - 10) = 30, but the percent of decrease will be calculated with respect to 40.

Therefore, the % decrease = \frac{30 \times 100}{40} = 75%.

Hence, Eduardo's second statement is wrong. (Answer)

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<img src="https://tex.z-dn.net/?f=f%28x%29%20-%20%5Cfrac%7Bx%5E%7B2%7D-4%20%7D%7Bx%5E%7B4%7D%20%2Bx%5E%7B3%7D%20-4x%5E%7B2%7D-4%
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a) The given function is

f(x)=\frac{x^2-4}{x^4+x^3-4x^2-4}

The domain refers to all values of x for which the function is defined.

The function is defined for

x^4+x^3-4x^2-4\ne0

This implies that;

x\ne -2.69,x\ne 1.83

b) The vertical asymptotes are x-values that makes the function undefined.

To find the vertical asymptote, equate the denominator to zero and solve for x.

x^4+x^3-4x^2-4=0

This implies that;

x= -2.69,x=1.83

c) The roots are the x-intercepts of the graph.

To find the roots, we equate the function to zero and solve for x.

\frac{x^2-4}{x^4+x^3-4x^2-4}=0

\Rightarrow x^2-4=0

x^2=4

x=\pm \sqrt{4}

x=\pm2

The roots are x=-2,x=2

d) The y-intercept is where the graph touches the y-axis.

To find the y-inter, we substitute;

x=0 into the function

f(0)=\frac{0^2-4}{0^4+0^3-4(0)^2-4}

f(0)=\frac{-4}{-4}=1

e) to find the horizontal asypmtote, we take limit to infinity

lim_{x\to \infty}\frac{x^2-4}{x^4+x^3-4x^2-4}=0

The horizontal asymtote is y=0

f) The greatest common divisor of both the numerator and the denominator is 1.

There is no common factor of the numerator and the denominator which is  at least a linear factor.

Therefore the function has no holes.

g) The given function is a proper rational function.

There is no oblique asymptote.

See attachment for graph.

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