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3 years ago

A sack of Nickels has a mass of 45 g. There are 9 nickels in the stack

Mathematics

2 answers:

NARA [144]3 years ago

7 0

To do this you will take the mass of the sack which is 45g.

Divide this amount by the number of nickels in the sack which is 9.

$45 \div 9 = 5$

So the mass of one nickel is 5g.

Tanya [424]3 years ago

3 0

Hello there!

9 nickels = 45g

1 nickel = ?

Then cross multiply that

45 * 1 = 9 * x

45 = 9x

x = 45/9

x = 5

As always, it is my pleasure helping students like you!

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goblinko [34]

Answer:

4th choice $\bold{f^{-1}(x) = \sqrt{x} + 4}$

Step-by-step explanation:

<u>Definition of the inverse of a function</u>

A function g is the inverse of a function f if whenever y=f(x) then x=g(y). In other words, applying f and then g is the same thing as doing nothing. We can write this in terms of the composition of f and g as g(f(x))=x. The domain of f becomes the range of g and the range of f becomes the domain of g

To solve for the inverse of the function $f(x) =\left(x-4\right)^2$

Let $y=\left(x-4\right)^2$

$\mathrm{Replace}\:x\:\mathrm{with}\:y$ $\text{ and replace }\:y\:\mathrm{with}\:x$

$x=\left(y-4\right)^2$

Switch sides
$\left(y-4\right)^2=x$

Take square roots on both sides
$y-4=\pm\sqrt{x}$

Add 4 on both sides to solve for y

$y = \pm\sqrt{x} + 4$

We have two solutions

$y=\sqrt{x}+4,\:y=-\sqrt{x}+4$

To determine which one of these to be chosen not that in the given choices we can eliminate the first two since x cannot be negative

The third choice can also be eliminated since
$-\sqrt{x} + 4$ is a decreasing function for $x \ge 0$

So the last answer choice is correct and the inverse of $f(x) = (x-4)^2$

is given by $f^{-1}(x) = \sqrt{x} + 4$

Answer:4th choice $\bold{f^{-1}(x) = \sqrt{x} + 4}$

Note

Domain of (x-4)² is [4, ∞) since x ≥ 4 and (x-4)² cannot be negative

Range of (x-4)² is [0, ∞)

Domain of $\sqrt{x}\:+\:4$ is [0, ∞)

Range of $\sqrt{x}\:+\:4$ is [4, ∞)

so indeed the domain of (x-4)² has become the range of $\sqrt{x}\:+\:4$ and the range of (x-4)² has become the domain of $\sqrt{x}\:+\:4$

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