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Len [333]
3 years ago
12

PLEASE HELP!!

Mathematics
2 answers:
cluponka [151]3 years ago
8 0

Answer:

x\in(-\infty,-2)\cup(5,\infty)

Step-by-step explanation:

The duadratic function g(x)=x^2+20 begin to exceed the linear function f(x)=3x+30 when g(x)>f(x)

Solve this inequality:

x^2+20>3x+30\\ \\x^2-3x+20-30>0\\ \\x^2-3x-10>0\\ \\x^2-5x+2x-10>0\\ \\x(x-5)+2(x-5)>0\\ \\(x-5)(x+2)>0

This inequality is equivalent to

\left[\begin{array}{l}\left\{\begin{array}{l}x-5>0\\x+2>0\end{array}\right.\\ \\\left\{\begin{array}{l}x-5-2\end{array}\right.\\ \\\left\{\begin{array}{l}x

Answer: x\in(-\infty,-2)\cup(5,\infty)

VMariaS [17]3 years ago
5 0

Answer:

x=5

Step-by-step explanation:

We are given that  

f(x)=3x+30

g(x)=x^2+20

We have to find the positive  integer value of x for which the quadratic function g(x) begin to exceed the linear function f(x).

g(x) > f(x)

x^2+20 > 3x+30

x^2+20-3x-30 >0

x^2-3x-10 > 0

(x-5)(x+2) > 0

x-5 > 0

x > 5

x+2 > 0

x >-2

Interval (5,\infty)

Therefore , g(x) exceed f(x) in the interval (5,\infty).

g(x) begin to exceed the linear function at x=5

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