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4 years ago

Evaluate this piecewise function

Mathematics

1 answer:

ohaa [14]4 years ago

7 0

Answer:

f(-2) = 2 and f(0) = 0.

Step-by-step explanation:

The piece wise function is given by, f(x) = x³ for x ≠ - 2 and f(x) = |x| for x = - 2.

Now, as defined above at x = -2, f(x) = |- 2| = 2

So, f(-2) = 2

Again, at x = 0, f(x) = x³ = (0)³ = 0.

So, f(0) = 0.

Hence, f(-2) = 2 and f(0) = 0. (Answer)

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The two legs of a right triangle measure 6 and 2. What is the length of the hyptenuse?

vitfil [10]

Answer:

6. 3 (rounded off to one decimal place)

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4 years ago

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Viktoriya solved an equation as shown below 5(x-2)-3x=2(x-3)-16 5x-10-3x=2x-6-16 2x-10=2x-22 -10=-22 She says that the equation

Anvisha [2.4K]

For this case we have the following equation:

$5 (x-2) -3x = 2 (x-3) -16$

The steps used by Viktoriya to solve are:

Step 1:

Distributive property:

$5x-10-3x = 2x-6-16$

Step 2:

Combine similar terms:

$2x-10 = 2x-22$

Step 3:

Simplify:

$-10 = -22$

Viktoriya's error is in the conclusion of the problem.

The equation has no solution, because:

Left side ≠ right side

Answer:

Viktoriya's error is: the equation has infinitely many solutions.

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4 years ago

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What answer is the area to 86ft

allsm [11]

Answer:

The only one would be 2 and 43

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3 years ago

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ABC ~ XYZ Find the length of BC (need an answer as soon as possible, with showing work please)

Snezhnost [94]

Answer:

BC is equal to 7

Step-by-step explanation:

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3 years ago

Which pair of funtions is not a pair of inverse functions? please help!!

antiseptic1488 [7]

Answer:

$f(x)=\frac{x}{x+20} , g(x)=\frac{20x}{x-1}$

Step-by-step explanation:

we know that

To find the inverse of a function, exchange variables x for y and y for x. Then clear the y-variable to get the inverse function.

we will proceed to verify each case to determine the solution of the problem

case A) $f(x)=\frac{x+1}{6} , g(x)=6x-1$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=\frac{y+1}{6}$

Isolate the variable y

$6x=y+1$

$y=6x-1$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=6x-1$

therefore

f(x) and g(x) are inverse functions

case B) $f(x)=\frac{x-4}{19} , g(x)=19x+4$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=\frac{y-4}{19}$

Isolate the variable y

$19x=y-4$

$y=19x+4$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=19x+4$

therefore

f(x) and g(x) are inverse functions

case C) $f(x)=x^{5}, g(x)=\sqrt[5]{x}$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=y^{5}$

Isolate the variable y

fifth root both members

$y=\sqrt[5]{x}$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=\sqrt[5]{x}$

therefore

f(x) and g(x) are inverse functions

case D) $f(x)=\frac{x}{x+20} , g(x)=\frac{20x}{x-1}$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=\frac{y}{y+20}$

Isolate the variable y

$x(y+20)=y$

$xy+20x=y$

$y-xy=20x$

$y(1-x)=20x$

$y=20x/(1-x)$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=20x/(1-x)$

$\frac{20x}{1-x}\neq \frac{20x}{x-1}$

therefore

f(x) and g(x) is not a pair of inverse functions

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