Rounding to the nearest tenth which of the following gives an overestimate a 37.63 times 0.54 b 54.49 times 0.45 c 1.19 times 1.45 d 0.81 times 8.01 e 27.83 times 13.64

Step-by-step explanation:

Rules of rounding off:

1.) If the digit following the last digit to be kept is less than five, the last digit is left unchanged.

2.) If the digit following the last digit to be kept is five or more, the last digit to be kept is increased by one.

A. 37.63 times 0.54

37.63 × 0.54  = 20.3202

20.3202 ≈ 20.3

B. 54.49 times 0.45

54.49 × 0.45  = 24.5205

24.5205 ≈  24.5

C. 1.19 times 1.45

1.19 × 1.45  = 1.7255

1.7255 ≈   1.7

D. 0.81 times 8.01

0.81 × 8.01  = 6.4881

6.4881 ≈ 6.5

E. 27.83 times 13.64

27.83 × 13.64 = 379.6012

379.6012 ≈ 379.6

pi x^2

step-by-step explanation:

pi times x squared

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.