Mathematics: Geometry: asap! show all work. incomplete or off-topic answers will be reported. 37. m is the midpont of line jk. find the coordinates of k. j(0,11), m(-3,2) 40. find the circumference and area for each circle in terms of pi. r = 3 in.

Geometry: asap! show all work. incomplete or off-topic

emily0981

11.03.2022

midpoint of (x1,y1) and (x2,y2) is $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}$

so if midpoint is (-3,2) and one point is (0,11) then

$(-3,2)=(\frac{0+x_2}{2}, \frac{11+y_2}{2})$

$-3=\frac{x_2}{2}$

-6=x_2

$2=\frac{11+y_2}{2}$

4=11+y_2

-7=y_2

the coordinates of k is (-6,-7)

circumference=2*pi*r

area=pi*r^2

if r=3 then

circumference=2*pi*3=6pi in

area=pi*3^2=pi*3*3=9pi in²

0,0(0 оценок)

jeffro198004

26.09.2020

$\huge \bf༆ Answer ༄$

Use PEDMAS/BODMAS rule for simplification !

Let's solve ~

$\sf{48} \div 6 + (5 \times 6) \times 13 + 6$ $\sf48 \div 6 + 30 \times 13 + 6$ $\sf8 + 30 \times 13 + 6$ $\sf8 + 390 + 6$ $\sf404$

0,0(0 оценок)