What is the midpoint of a segment whose endpoints are (3, 7) and (7, 3)?

2 answers:

7 0

Midpoint of a segment whose endpoints are (x₁,y₁) and (x₂,y₂)
M((x₁+x₂)/2 , (y₁+y₂)/2)
1)
(3,7)
(7,3)

M( (3+7) /2 , (7+3)/2 )
M(10/2 , 10/2)
M(5,5)

B(5,5).

2)
Distance between the points (x₁,y₁) and (x₂,y₂)

Distance=√[(x₂-x₁)²+(y₂-y₁)₂]
(6,32)
(-8,-16)
distance=√[(-8-6)²+(-16-32)²]
d=√[(-14)²+(-48)²]
d=√(196+2304)
d=√2500
d=50

D. 50

(2,3)
(7,4)
d=√[(7-2)²+(4-3)²]
d=√(5²+1²)
d=√(25+1)
d=√26≈5.1

d≈5.1

Ivan3 years ago

3 0

Answer:

The answer to the third question is 10..

Step-by-step explanation:

You might be interested in

Find the volume of the composite solid shown below, Round answer to the nearest hundredth.

MrMuchimi

First you’d start off by finding the volume of the rectangular prism on the bottom. The formula for this is lwh or (length*width*height) so you’d plug in the numbers and get 4*4*5 which multiplies out to be 80inches cubed. Next you’d have to find the volume of the pyramid by using the formula lwh/3 or ((length*width*height)/3) so it would be ((4*4*6)/3) and that equals 32inches cubed. Finally, you add the 2 volumes together so 80inches cubed+32inches cubed and get 112inches cubed. Hope this helps

4 0

2 years ago

If DK PVX, which of the following statements must also be true Check all that apply

Oxana [17]

I think it’s AVPXADK

8 0

3 years ago

The original purchase price of a car is 16,000, Each year, its value depreciates by 10%. Three years after its purchase, what is

Tju [1.3M]

The answer is 11,664

3 0

3 years ago

Rex, Paulo, and Ben are standing on shore watching for dolphins. Paulo sees one surface directly in front of him about a hundred

Ksivusya [100]

1. $m\angle BAC=m\angle CAD,\ m\angle ACB=m\angle ADC=90^{\circ}$ , then $m\angle ABC=m\angle ACD$ and triangles ADC and ACB are similar by AAA theorem.

2. The ratio of the corresponding sides of similar triangles is constant, so

$\dfrac{AC}{AB}= \dfrac{AD}{AC}$ .

3. Knowing lengths you could state that $\dfrac{b}{c}= \dfrac{e}{b}$ .

4. This ratio is equivalent to $b^2=ce$ .

5. $m\angle ABC=m\angle CBD,\ m\angle ACB=m\angle CDB=90^{\circ}$ , then $m\angle BAC=m\angle BCD$ and triangles BDC and BCA are similar by AAA theorem.