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3 years ago

Find a distance between the two points

Mathematics

2 answers:

Alekssandra [29.7K]3 years ago

6 0

i dont seem to understamd

yanalaym [24]3 years ago

4 0

Step-by-step explanation:

To find the distance you need the distance formula. The distance formula is $\sqrt{(x_{2}-x_{1} )^{2} + (y_{2}-y_{1})^2 }$ . Order the pairs with a 1 or a 2. Look at he point that you labeled 2. Substitute any of the points with a 2 subscript into the equation. The x value is the first number in the ordered pair and the y value is the second number in the ordered pair. Repeat this with the ordered pair you labeled as 1. Substitute the numbers in. You should get something like this: $\sqrt{(-7-5)^{2} + (-7-9)^{2} }$ . If you are doing this without a calculator, it is easier than you think. Remember PEMDAS? Well apply that with this. Find out the solution to the parenthesis first. The first parenthesis should come out with $(-12)^{2}$ . Finish the expression off by evaluating the expontent. That comes out to 144. Try this with the next parenthesis... The answer should be 256. Your square root should look like this: $\sqrt{144+256}$ . Add the numbers under the parenthesis together and you should get $\sqrt{400}$ . I'm sure you have a calculator in you hand ready to solve this. NOW YOU FIND THE ANSWER!!!

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How many solutions are there to the equation below?<br> 6x + 33 = 6(x+6)

Ainat [17]

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

Let's solve ~

$6x + 33 = 6(x + 6)$

$6x + 33 = 6x + 36$

Here, 6x is present on both the sides of equation so they cancel out ~ therefore there is no one solution satisfying the equation ~

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2 years ago

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uysha [10]

Step-by-step explanation:

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2 years ago

8/2(2+2)=?<br><br> 100 points to the best answer!

Effectus [21]

Answer:

Step-by-step explanation:

8/2(2+2)

- According to the rules of BEDMAS, solve the values inside the bracket.

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2 years ago

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Determine whether the equation x^3 - 3x + 8 = 0 has any real root in the interval [0, 1]. Justify your answer.

nikdorinn [45]

Answer:

The equation does not have a real root in the interval $\rm [0,1]$

Step-by-step explanation:

We can make use of the intermediate value theorem.

The theorem states that if $f$ is a continuous function whose domain is the interval [a, b], then it takes on any value between f(a) and f(b) at some point within the interval. There are two corollaries:

If a continuous function has values of opposite sign inside an interval, then it has a root in that interval. This is also known as Bolzano's theorem.
The image of a continuous function over an interval is itself an interval.

Of course, in our case, we will make use of the first one.

First, we need to proof that our function is continues in $\rm [0,1]$ , which it is since every polynomial is a continuous function on the entire line of real numbers. Then, we can apply the first corollary to the interval $\rm [0,1]$ , which means to evaluate the equation in 0 and 1:

$f(x)=x^3-3x+8\\f(0)=8\\f(1)=6$

Since both values have the same sign, positive in this case, we can say that by virtue of the first corollary of the intermediate value theorem the equation does not have a real root in the interval $\rm [0,1]$ . I attached a plot of the equation in the interval $\rm [-2,2]$ where you can clearly observe how the graph does not cross the x-axis in the interval.