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3 years ago

5

What is the domain of the function y=10-5x

2 answers:

sergey [27]3 years ago

8 0

Any real number (-inf, inf)

vampirchik [111]3 years ago

8 0

Domain of definition of a function is the set of numbers which the variable attains and for which the function is defined.

f(x) = sqrt (5x-10)

here x can have value equal to any real number >=2 because if x attains value less that 2, 5x-10 becomes negative and sqrt(5x-10) has no real value.

therefore the domain of the function f(x) is (2, infinity) inclusive of 2.

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3 years ago

The function f(x)=2/3x-4 was replaced with f(x+k)=2/3x-6 what is the value of k

HACTEHA [7]

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The given function $f(x)=\frac{2x}{3}-4$ was replaced with $f(x+k)=\frac{2x}{3}-6$ then the value of k is -3

that is k=-3.

Step-by-step explanation:

The given function is $f(x)=\frac{2x}{3}-4\hfill (1)$

and when x=x+k then $f(x+k)=\frac{2x}{3}-6\hfill (2)$

Now to find the value of k:

Put x=x+k in the function f(x)

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

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

$f(x+k)=\frac{2x+2k}{3}-4\hfill (3)$

Now comparing equations (2) and (3) we get

$f(x+k)=\frac{2x+2k}{3}-4=\frac{2x}{3}-6$

$\frac{2x+2k}{3}-4=\frac{2x}{3}-6$

$\frac{2x+2k}{3}-4-\frac{2x}{3}+6=0$

$\frac{2x}{3}+\frac{2k}{3}-4-\frac{2x}{3}+6=0$

$\frac{2k}{3}+2=0$

$\frac{2k}{3}=-2$

$k=\frac{-2\times 3}{2}$

$k=-3$

Therefore k=-3

Therefore the given function $f(x)=\frac{2x}{3}-4$ was replaced with $f(x+k)=\frac{2x}{3}-6$ then the value of k is -3

that is k=-3.

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2 years ago