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3 years ago

What is the constant of posprtainality in the equation y=0.41

Mathematics

1 answer:

KATRIN_1 [288]3 years ago

5 0

Answer:

The constant of proportionality in the equation y=0.41 is 0.41

Step-by-step explanation:

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The graph of a system of parallel lines will have no solutions. (1 point) Select one: a. Always b. Sometimes c. Never

Andreas93 [3]

Answer:

Always

Step-by-step explanation:

They never intersect

8 0

3 years ago

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For which function is f(x) equal to f^1(x) ( answer choices in picture )

Sonja [21]

Answer:

C. $f(x)=\frac{x+1}{x-1}$

Step-by-step explanation:

Let's find the inverse of each of the given options.

Option A:

$f(x)=\frac{x+6}{x-6}\\y=\frac{x+6}{x-6}$

To find $f^{-1}(x)$ , replace 'x' with 'y' and 'y' with 'x'. This gives,

$x=\frac{y+6}{y-6}$

Rewrite in terms of 'y'. This gives,

$x(y-6)=y+6\\xy-6x=y+6\\xy-y=6x+6\\y=\frac{6x+6}{x-1}$

The given function $y=\frac{6x+6}{x-1}\ne y=\frac{x+6}{x-6}$

So, option A is incorrect.

Option B:

$f(x)=\frac{x+2}{x-2}\\y=\frac{x+2}{x-2}$

To find $f^{-1}(x)$ , replace 'x' with 'y' and 'y' with 'x'. This gives,

$x=\frac{y+2}{y-2}$

Rewrite in terms of 'y'. This gives,

$x(y-2)=y+2\\xy-2x=y+2\\xy-y=2x+2\\y=\frac{2x+2}{x-1}$

The given function $y=\frac{2x+2}{x-1}\ne y=\frac{x+2}{x-2}$

So, option B is incorrect.

Option C:

$f(x)=\frac{x+1}{x-1}\\y=\frac{x+1}{x-1}$

To find $f^{-1}(x)$ , replace 'x' with 'y' and 'y' with 'x'. This gives,

$x=\frac{y+1}{y-1}$

Rewrite in terms of 'y'. This gives,

$x(y-1)=y+1\\xy-x=y+1\\xy-y=x+1\\y=\frac{x+1}{x-1}$

The given function $y=\frac{x+1}{x-1}\ equals\ y=\frac{x+1}{x-1}$

So, option C is correct.

Option D:

$f(x)=\frac{x+5}{x-5}\\y=\frac{x+5}{x-5}$

To find $f^{-1}(x)$ , replace 'x' with 'y' and 'y' with 'x'. This gives,

$x=\frac{y+5}{y-5}$

Rewrite in terms of 'y'. This gives,

$x(y-5)=y+5\\xy-5x=y+5\\xy-y=5x+5\\y=\frac{5x+5}{x-1}$

The given function $y=\frac{5x+5}{x-1}\ne y=\frac{x+6}{x-6}$

So, option D is incorrect.

Therefore, only option C is correct.

7 0

3 years ago

Why is adding 3/9 and 6/9 different from multpying the two fractions? Explain

tankabanditka [31]

<h3>♫ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ♫</h3>

➷ When you add fractions, you have to have the same denominator whereas, when you multiply them, you don't have to. (In this case the denominators are already the same.) Also, multiplying the fractions would make the result smaller whereas adding them would make it larger.

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

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4 years ago

Simplify this expression

Gnesinka [82]

It’s A!! hope that helped

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3 years ago

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Do you know the answer to this question

kozerog [31]

The correct answer is: [C]: " f(x) = log₃ (x - 3) " .
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3 years ago