Find the midpoint of the segment with the following endpoints. (5,9) and (0,3)

Mathematics

2 answers:

Mnenie [13.5K]3 years ago

8 0

Answer:

(5/2, 6)

Step-by-step explanation:

(x, y)=[(5+0)/2, (9+3)/2]

(x, y)=[5/2, 12/2]

(x, y)=(5/2, 6)

miss Akunina [59]3 years ago

3 0

Answer:

( 2.5, 3 )

Step-by-step explanation:

Assuming AC is the line and B is the mid-point; knowing ( 5, 9 ) and ( 0, 3 ) are the endpoints, you could add a number by itself to see what would equal 5 and do the same thing for 9.

2.5 + 2.5 = 5

3 + 3 = 6 + 3 = 9

Making the mid-point of the segment ( 2.5, 3 )

You might be interested in

Someone please help me out I rlly need help

Anit [1.1K]

The equation that could be used to solve for x is (3x-1) = 140 and the value of x is 47°.

According to the question,

We have the following information:

We have two lines that are intersecting each other at a point. The angles made on this point are given.

Now, we know that these two angles are vertically opposite angles. And we also know that vertically opposite angles are equal.

So, we will have the following equation:

3x-1 = 140

Now, we can easily find the value of x from this equation:

Adding 1 on both sides of the equation:

3x = 140+1

3x = 141

Dividing by 3 on both sides:

x = 141/3

x = 47°

Hence, the equation that could be used to solve for x is (3x-1) = 140 and the value of x is 47°.

To know more about equation here

brainly.com/question/18067015

#SPJ1

5 0

11 months ago

In a large population of adults, the mean IQ is 112 with a standard deviation of 20. Suppose 200 adults are randomly selected fo

Snezhnost [94]

Answer:

(c)approximately Normal, mean 112, standard deviation 1.414.

Step-by-step explanation:

To solve this problem, we have to understand the Central Limit Theorem

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .