Which represents the solution set of 5(x+5)<85?

2 answers:

8 0

Isolate the VARIABLE by dividing each side by FACTROS that don't contain the VARIABLE .
YOUR answer is x<12

True [87]3 years ago

8 0

The correct answer is: [A]: " x < 12 " .
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Given:
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" 5(x + 5) < 85 " ; Simplify/ solve for "x" ;
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Method 1)
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Note the "distributive property of multiplication" :
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a(b + c) = ab + ac ;

a(b – c) = ab – ac .
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So, take the "left-hand side" of the inequality; and expand:
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 → 5(x + 5) = (5*x) + (5*5) = 5x + 25 ;

Now, rewrite the inequality:

 → 5x + 25 < 85 ;

Subtract "25" from each side of the inequality ;

 → 5x + 25 – 25 < 85 – 25 ;

to get:
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 → 5x < 60 ;

Divide each side of the inequality by "5" ;

 → 5x / 5 < 60 / 5 ;

 → x < 12 ; which is: Answer choice: [A]: " x < 12 " .
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Method 2)
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Given:
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" 5(x + 5) < 85 " ; Simplify/ solve for "x" ;
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Divide EACH SIDE of the equation by "5" ;

[5(x + 5) ] / 5 < 85 / 5 ;

to get:

 → x + 5 < 17 ;

Subtract "5" from EACH SIDE of the inequality:

→ x + 5 – 5 < 17 – 5 ;

to get:

→ x < 12 ; which is: Answer choice: [A]: " x < 12 " .
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The function f(x)=(x-5)^2 +2 is not one-to-one. Identify a restricted domain that makes the function one-to-one, and find the in

Anon25 [30]

We have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

We know that vertex of this quadratic function occurs at (5,2).

Further, we know that range of this function is $[2,\infty)$ .

If we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

Let us know find its inverse.

$y=(x-5)^{2}+2$

Upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

Let us now solve this function for y.

$(y-5)^{2}=x-2\\
y-5=\pm \sqrt{x-2}\\
y=5\pm \sqrt{x-2}\\$

Hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .