Identify the asymptotes of the graph.

Answer: the amount of dimes is $1.15 this is also equivalent to 11 dimes and an extra nickel.

Explanation:

80 nickels =$4.00
$5.15-$4.00= $1.15
11 dimes= $1.10
1 nickel= $0.05
$1.10+$0.05= $1.15

hope this helps! :)

5 0

3 years ago

Which statement is true about the function f(x)= StartRoot negative x EndRoot?

UNO [17]

Answer:

The domain of the graph is all real numbers less than or equal to 0

Step-by-step explanation:

we have the function

$f(x)=\sqrt{-x}$

we know that

The radicand must be positive

so

$-x \geq 0$

Solve for x

Multiply by -1 both sides

$x\leq 0$

The domain is the interval ------> (-∞,0]

All real numbers less than or equal to zero

The range of the function is the interval -----> [0,∞)

$f(x)\geq 0$

see the attached figure to better understand the problem

All real numbers greater than or equal to zero

<em><u>Verify each statement</u></em>

case 1) The domain of the graph is all real numbers

The statement is false

Because, the domain is all real numbers less than or equal to zero

case 2) The range of the graph is all real numbers

The statement is false

Because, the range is all real numbers greater than or equal to zero

case 3) The domain of the graph is all real numbers less than or equal to 0

The statement is true

case 4) The range of the graph is all real numbers less than or equal to 0

The statement is false

Because, the range is all real numbers greater than or equal to zero

4 0

3 years ago

Read 2 more answers

Please help me answer this question

ratelena [41]

By the definition of the hyperbolic function tanh x, we have proven that $\frac{1-tanh \ x}{1 + tanh \ x}=e^{-2x}$

<h3>Hyperbolic functions & Proof of identities </h3>

By definition

$tanh \ x=\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}}$

Then,

$\frac{1-tanh \ x}{1 + tanh \ x}=\frac{1-\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} }{1+ \frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} }$

$=1-\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} \div (1+ \frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} })$

$=\frac{e^{x} +e^{-x}-(e^{x} -e^{-x})}{e^{x} +e^{-x}} \div (\frac{e^{x} +e^{-x}+e^{x} -e^{-x}}{e^{x} +e^{-x}} })$

$=\frac{e^{x} +e^{-x}-e^{x} +e^{-x}}{e^{x} +e^{-x}} \div \frac{e^{x} +e^{-x}+e^{x} -e^{-x}}{e^{x} +e^{-x}} }$

$=\frac{e^{-x}+e^{-x}}{e^{x} +e^{-x}} \div \frac{e^{x} +e^{x} }{e^{x} +e^{-x}} }$

$=\frac{2e^{-x}}{e^{x} +e^{-x}} \div \frac{2e^{x} }{e^{x} +e^{-x}} }$

$=\frac{2e^{-x}}{e^{x} +e^{-x}} \times \frac{e^{x} +e^{-x}}{2e^{x}}$

$=\frac{2e^{-x}}{1} \times \frac{1}{2e^{x}}$

$=\frac{2e^{-x}}{2e^{x}}$

$=\frac{e^{-x}}{e^{x}}$
$=e^{-x} \times \frac{1}{e^{x}}$

$= e^{-x} \times e^{-x}$

$= e^{-x+-x}$

$= e^{-x-x}$

$= e^{-2x}$

Hence, we have proven that $\frac{1-tanh \ x}{1 + tanh \ x}=e^{-2x}$

Learn more on Proof of Identities here: brainly.com/question/2561079

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6 0

2 years ago