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natima [27]
3 years ago
13

Find the slope between (3,2) and (-2,3)

Mathematics
1 answer:
blagie [28]3 years ago
3 0
(x_1;\ y_1);\ (x_2;\ y_2)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}\\\\(3;\ 2);\ (-2;\ 3)\\\\\boxed{m=\dfrac{3-2}{-2-3}=\dfrac{1}{-5}=-\dfrac{1}{5}}
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Answer:

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3 years ago
How do I estimate the square root and round to the nearest integer?
Troyanec [42]

The square root of a number y is a number x such that x^2=y

So, if a number is a perfect square, its square root is an integer. For example, the square root of 25 is 5, because 5^2=25.

If a number is not a perfect square, its square root is not an integer, which means that it is an irrational number, included between two integers.

If you want to find these two integers, you need to find two (consecutive) integers such that the first is "not enough", whereas the second is "too much".

Here's an example, suppose we want to find the square root of 20. We try some integers:

\left.\begin{array}{cc}1^2 = 1 & \text{Not Enough}\\2^2 = 4 & \text{Not Enough}\\3^2 = 9 & \text{Not Enough}\\4^2 = 16 & \text{Not Enough}\\5^2 =25 & \text{Too Much}\end{array}\right.

So, when we square 4 is slightly less than 20, and when we square 5 is slightly more than 20. This means that there must exist a number between 4 and 5 that, when squared, is exactly 20.

So, for sure 4 < \sqrt{20} < 5

In order to round to the nearest integer, we consider that 4^2=16 was 4 units away from the goal (20), while 5^2=25 was 5 units away from the goal. So, 4^2 is nearest to the goal than 5^2, and so we round

\sqrt{20}\approx 4

For the record, we have

\sqrt{20} \approx 4.472135955\ldots

so our approximation was correct.

3 0
3 years ago
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A perpendicular line should be in the same plane and intersect each other.
7 0
4 years ago
Read 2 more answers
The difference of Katie's age and<br> negative two is fourteen. How old is<br> Katie?
DiKsa [7]

Answer:

12?

¯\_(ツ)_/¯

Step-by-step explanation:

8 0
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