What is the approximate area of a circle whose diameter is 6.5 m? Use 3 14 for T. Enter your answer in the box. Round your final answer to two decimal places​

yes

step-by-step explanation:

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.

the anwser is x< -17/2

here is me explaining it:

the pizza would be delivered around 5, give or take 15 minutes. this means it could be delivered anywhere from 4: 45 to 5: 15

let's say 5: 00 is equivalent to 0 on a number line and x represents time. 4: 45 would be equal to -15x because we are 15 minutes behind our starting point, 5: 00. 5: 15 would be equal to 15x because we are 15 minutes ahead of your starting point (which means you're late on the pizza delivery! )

absolute value is very confusing. the basic definition is how far you are from your starting point. if we deliver the pizza at 4: 45 (15 minutes before the starting point) we would be -15. this means we are 15 away from our starting point, 5: 00. that's the absolute value. |-15| (absolute value of negative fifteen) is equal to 15. if you did not understand this i suggest rereading this paragraph.

the normal way you would write this scenario would be 4.75 (4 and 3 quarters, or 4: 45. you can also write as a fraction.) is less than x and x is less than 5.25 (5 and 1 quarter or 5: 15). 4.75< x< 5.25 the reason why i am using numbers is because you cannot represent typical time as values on a graph, number line, or in an equation.

another way to write this scenario is with absolute value. we can write this as |x| being less than or equal to .25 (remember, this just means 15 minutes).           |x| ≤ .25.

sorry if the end was confusing. because we are talking about time this whole question is 10x harder. like i said earlier, you cannot represent time as, well, time, in a graph, equation, or number line. i set the values as follows: 1/4 or .25 for 15 minutes, and 1 for 1 hour. this is just a common way to set up graphs involving time.

|x| ≤ .25

i hope this ! : )

tell me if you need more assistance