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Mars2501 [29]
3 years ago
12

Which elements in the set below are integers? –3, 3.7, √ 81 , –7.34, √ 8 , 5, 56⁄7, –1 A. –3, √ 81 , 5, 56⁄7, –1 B. 3.7,√ 81 , √

8 , 5, 56⁄7 C. –3, 3.7, –7.34, –1 D. 3.7, –7.34, √ 8
Mathematics
2 answers:
Doss [256]3 years ago
4 0

The answer is:  A. –3, \sqrt{81}, 5, \frac{56}{7}.

The explanation is shown below:

1) By definition, the integers include whole numbers, negative numbers and zero.

2) Therefore, keeping this information on mind, you have that:

-3 is a negative number without decimal part, therefore, it is an integer.

3.7 is decimal number, so it is not an integer.

\sqrt{81}=9 (It is a whole number, therefore, it is an integer).

-7.34 is a decimal number, so it is not an integer.

\sqrt{8}=2.828 is an irrational number. It is not an integer.

5 is a whole number, therefore, it is a integer.

\frac{56}{7}=8 (It is a whole number, then, it is an integer).

Aleks04 [339]3 years ago
3 0

Answer:

A

Step-by-step explanation:

–3, √ 81 , 5, 56⁄7

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Write the equation of the line that passes through (−3,1) and (2,−1) in slope-intercept form
Alex787 [66]

Answer:

y=-\frac{2}{5}x-\frac{1}{5}

Step-by-step explanation:

The equation of a line is y = mx + b

Where:

  • m is the slope
  • b is the y-intercept

First, let's find what m is, the slope of the line.

Let's call the first point you gave, (-3,1), point #1, so the x and y numbers given will be called x1 and y1.

Also, let's call the second point you gave, (2,-1), point #2, so the x and y numbers here will be called x2 and y2.

Now, just plug the numbers into the formula for m above, like this:

m = -\frac{2}{5}

So, we have the first piece to finding the equation of this line, and we can fill it into y=mx+b like this:

y=-\frac{2}{5}x + b

Now, what about b, the y-intercept?

To find b, think about what your (x,y) points mean:

  • (-3,1). When x of the line is -3, y of the line must be 1.
  • (2,-1). When x of the line is 2, y of the line must be -1.

Now, look at our line's equation so far: y=-\frac{2}{5}x + b. b is what we want, the --\frac{2}{5} is already set and x and y are just two 'free variables' sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the two points (-3,1) and (2,-1).

So, why not plug in for x and y from one of our (x,y) points that we know the line passes through? This will allow us to solve for b for the particular line that passes through the two points you gave!

You can use either (x,y) point you want. The answer will be the same:

  • (-3,1). y = mx + b or 1=-\frac{2}{5} * -3 + b, or solving for b: b = 1-(-\frac{2}{5})(-3).b = -\frac{1}{5}.
  • (2,-1). y = mx + b or -1=-\frac{2}{5} * 2 + b, or solving for b: b = 1-(-\frac{2}{5})(2). b = -\frac{1}{5}.

See! In both cases, we got the same value for b. And this completes our problem.

The equation of the line that passes through the points  (-3,1) and (2,-1) is y=-\frac{2}{5}x-\frac{1}{5}

8 0
3 years ago
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