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sveticcg [70]
3 years ago
13

What kind of number is -5? Is it rational or something like that?

Mathematics
2 answers:
Arte-miy333 [17]3 years ago
6 0
ANSWER
-5 is a real number, a rational number, and an integer

EXPLANATION
Rational number are numbers which you can write as a quotient of two integers. Integers are non-decimal numbers that can be positive or negative. We can therefore write -5 as -5 divided by 1 (-5/1) and thus classify -5 as rational.

In fact, you can classify -5 into other categories.

-5 is all of the following
⇒ it is a real number
⇒ it is a rational number
⇒ it is an integer

Whole numbers can be considered as {0, 1, 2, 3, ...}
Natural numbers can be considered as {1,2,3,...}

so -5 is not a whole or natural number.

\boxed{ \begin{array}{l r} \textbf{real numbers } (\mathbb{R}) \\ \boxed{ \begin{array}{l} \textbf{rational numbers } (\mathbb{Q}) \\ \boxed{ \begin{array}{l} \textbf{integers } (\mathbb{Z})\\ \boxed{ \begin{array}{l} \text{\footnotesize whole numbers ($\mathbb{W}$)} \\ \boxed{ \begin{array}{l} \text{\scriptsize natural numbers ($\mathbb{N}$)} \end{array}} \end{array}} \end{array}} \end{array}} & \boxed{ \begin{array}{c} \text{irrational numbers} \\ \ \\ \ \\ \ \\ \ \end{array}} \end{array}}
babymother [125]3 years ago
4 0
The number -5 is anything but natural number.
It is an integer (Z) ,a rational number (Q) and a real number (R) .

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A dog walker charges a flat rate of $6 per walk plus an hourly rate of $30. How much does the dog walker charge for a 45 minute
zvonat [6]

Step 1

<u>Find the equation in function notation</u>

Let

h-------> the number of hours

y-------> the function for the walker fee in dollars

we know that

the hourly rate is 30\frac{\$}{hour}

y=6+30h

in this linear equation

the independent variable is the variable h

the dependent variable is the variable y

<u>Convert to function notation</u>

Let

f(h)=y

f(h)=6+30h

Step 2

Find how much does the dog walker charge for a 45 minute walk

Convert the time in hours

1\ hour=60\ minutes

45\ minutes=45/60=0.75\ hours

substitute in the equation

For h=0.75\ hours

f(0.75)=6+30*(0.75)=\$28.5

therefore

<u>the answer is</u>

the dog walker charge for a 45 minute walk \$28.5

<u>Statements</u>

Step 3

<u>The dependent variable is the number of hours true or false</u>

The statement is false,

because the independent variable is the number of hours and the dependent variable is the walker fee in dollars

Step 4

<u>The function for the walker fee is f(h)= 30h+6 true or false</u>

The statement is True --------> see the Step 1

Step 5

<u>The dog walker charges $22.5 for a 45 min walk true or false</u>

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5 0
3 years ago
CALCULUS - Find the values of in the interval (0,2pi) where the tangent line to the graph of y = sinxcosx is
Rufina [12.5K]

Answer:

\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}

Step-by-step explanation:

We want to find the values between the interval (0, 2π) where the tangent line to the graph of y=sin(x)cos(x) is horizontal.

Since the tangent line is horizontal, this means that our derivative at those points are 0.

So, first, let's find the derivative of our function.

y=\sin(x)\cos(x)

Take the derivative of both sides with respect to x:

\frac{d}{dx}[y]=\frac{d}{dx}[\sin(x)\cos(x)]

We need to use the product rule:

(uv)'=u'v+uv'

So, differentiate:

y'=\frac{d}{dx}[\sin(x)]\cos(x)+\sin(x)\frac{d}{dx}[\cos(x)]

Evaluate:

y'=(\cos(x))(\cos(x))+\sin(x)(-\sin(x))

Simplify:

y'=\cos^2(x)-\sin^2(x)

Since our tangent line is horizontal, the slope is 0. So, substitute 0 for y':

0=\cos^2(x)-\sin^2(x)

Now, let's solve for x. First, we can use the difference of two squares to obtain:

0=(\cos(x)-\sin(x))(\cos(x)+\sin(x))

Zero Product Property:

0=\cos(x)-\sin(x)\text{ or } 0=\cos(x)+\sin(x)

Solve for each case.

Case 1:

0=\cos(x)-\sin(x)

Add sin(x) to both sides:

\cos(x)=\sin(x)

To solve this, we can use the unit circle.

Recall at what points cosine equals sine.

This only happens twice: at π/4 (45°) and at 5π/4 (225°).

At both of these points, both cosine and sine equals √2/2 and -√2/2.

And between the intervals 0 and 2π, these are the only two times that happens.

Case II:

We have:

0=\cos(x)+\sin(x)

Subtract sine from both sides:

\cos(x)=-\sin(x)

Again, we can use the unit circle. Recall when cosine is the opposite of sine.

Like the previous one, this also happens at the 45°. However, this times, it happens at 3π/4 and 7π/4.

At 3π/4, cosine is -√2/2, and sine is √2/2. If we divide by a negative, we will see that cos(x)=-sin(x).

At 7π/4, cosine is √2/2, and sine is -√2/2, thus making our equation true.

Therefore, our solution set is:

\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}

And we're done!

Edit: Small Mistake :)

5 0
2 years ago
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