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3 years ago

Rational numbers are _____ natural numbers.

Mathematics

2 answers:

zhannawk [14.2K]3 years ago

4 0

Rational numbers are sometimes natural numbers.

Harrizon [31]3 years ago

4 0

Answer:

sometimessssssss

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How do I use intercepts to graph 3y= - 5x - 30

expeople1 [14]

Answer:

y-intercept is (0,-10) and x-intercept is (-6,0). Connect them by a straight line to graph the given equation.

Step-by-step explanation:

The given equation of line is

$3y=-5x-30$

For x=0,

$3y=-5(0)-30$

$3y=-30$

$y=-10$

So, y-intercept is at point (0,-10).

For y=0,

$3(0)=-5x-30$

$0=-5x-30$

$5x=-30$

$x=-6$

So, x-intercept is at point (-6,0).

Now, plot the point (0,-10) and (-6,0) on a coordinate plane and connect them by a straight line to graph the given line as shown below.

7 0

3 years ago

Which graph does NOT represent a function?<br> R

vredina [299]

Answer:

The graph of the circle

Step-by-step explanation:

A function is defined to have only one output per input. Conducting the horizontal line test, we see that there are instances where the line intersects two points, which shows that some inputs have multiple outputs.

This does not fit the definition of a function.

I hope this helps! :)

5 0

3 years ago

Multiple-Choice Integration, Picture Included, Please Include Work

ValentinkaMS [17]

$\displaystyle\int_0^2\sqrt{4-x^2}\,\mathrm dx$

Recall that a circle of radius 2 centered at the origin has equation

$x^2+y^2=4\implies y=\pm\sqrt{4-x^2}$

where the positive root gives the top half of the circle in the x-y plane. The definite integral corresponds to the area of the right half of this top half. Since the area of a circle with radius $r$ is $\pi r^2$ , it follows that the area of a quarter-circle would be $\dfrac{\pi r^2}4$ .

You have $r=2$ , so the definite integral is equal to $\dfrac{2^2\pi}4=\pi$ .

Another way to verify this is to actually compute the integral. Let $x=2\sin u$ , so that $\mathrm dx=2\cos u\,\mathrm du$ . Now

$\displaystyle\int_0^2\sqrt{4-x^2}\,\mathrm dx=\int_0^{\pi/2}\sqrt{4-(2\sin u)^2}(2\cos u)\,\mathrm du=4\int_0^{\pi/2}\cos^2u\,\mathrm du$

Recall the half-angle identity for cosine:

$\cos^2u=\dfrac{1+\cos2u}2$

This means the integral is equivalent to

$\displaystyle2\int_0^{\pi/2}(1+\cos 2u)\,\mathrm du=2u+\sin2u\bigg|_{u=0}^{u=\pi/2}=\pi$

4 0

3 years ago

At the chocolatier, classic chocolate fudge costs 10 dollars a pound. Anna purchases 7/9 of a pound of the fudge. The cash regis

Elina [12.6K]

Anna pays $7.78. Or not rounded 7.777777..... repeated.

Hope this helps!

8 0

3 years ago

How do I write Domain and Range in inequality notation?

ArbitrLikvidat [17]

Answer:

Domain: (-infinity, infinity) Range: (-infinity, infinity)

Step-by-step explanation:

They are parabolas, therefore you can assume that they go on infinitely. To find range, you must look at your y values. Look for your lowest point. Because the line goes done forever, your beginning mark would be (-infinity.

To find the other part, you look at your positive y values. Look for the highest value. Because this goes on infinitely, the completed version of your notation would be (-infinity, infinity). Be sure to use the infinity symbol though, which looks like an 8 rotated 90 degrees.

To find domain, look at your x values. To begin, look at your left-most values, which would be the negative numbers. Because the line goes on forever to the left, your notation would be (-infinity. To find the other part of domain, look at your positive x values. Because this line goes on infinitely as well, the completed version of your notation would be (-infinity, infinity). Infinity is never bracketed, it is always in parenthesis.

8 0

3 years ago