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3 years ago

Evaluate f(x) when x=-10

Mathematics

1 answer:

adell [148]3 years ago

6 0

Answer:

f(x) = 10(10)+3 = 103 when x = 10

Step-by-step explanation:

Given a function piecewise as

f(x) = 2x-4 if -3<x<10

= x^2+3, if 10<=x<13

We see from the definition of f(x) that the function has domain as (-3,13) and two sets defined from (-3,10) and [10,13).

Since 10 less than or equal to sign is there in the second function interval, we find that 10 is included in that.

Hence for finding f(10) we use the function defined for interval including 10 i.e. x^2+3

Substitute 10 for x in x^2+3

f(10) = 10^2+3

=103

Hence answer is 103

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3 years ago

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Answer:

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3 years ago

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3 years ago

If 9 times a number is added to 18, the result is -27.<br> Find the number.

denis23 [38]

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3 years ago

Solve for the sqrt(x-2)+8=x

kirza4 [7]

<h3>Answer: x = 11</h3>

==============================================

Work Shown:

$\sqrt{x-2}+8=x\\\sqrt{x-2}=x-8\\\left(\sqrt{x-2}\right)^2=(x-8)^2 \text{ square both sides}\\x-2=x^2-16x+64\\0=x^2-16x+64-x+2\\0=x^2-17x+66\\x^2-17x+66 = 0\\(x-11)(x-6) = 0\\x-11 = 0 \text{ or } x-6 = 0\\x = 11 \text{ or } x = 6\\$

Those are the two possible solutions . We need to check each possible solution.

Plug in x = 11 then simplify. If we get the same number on both sides, then x = 11 is confirmed.

$\sqrt{x-2}+8=x\\\sqrt{11-2}+8=11\\\sqrt{9}+8=11\\3+8=11\\11=11\\$

We get 11 on both sides, so the solution x = 11 is confirmed.

Repeat for x = 6 as well

$\sqrt{x-2}+8=x\\\sqrt{6-2}+8=6\\\sqrt{4}+8=6\\2+8=6\\10=6\\$

We do not get the same thing on both sides, so x = 6 is not a solution. We consider this extraneous.

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3 years ago