Need a little . lines cd and de are tangent to circle a shown below: lines cd and de are tangent to circle a and intersect at point d. arc ce measures 126 degrees. point b lies on circle a. if arc ce is 126°, what is the measure of ∠cde? 126° 63° 117° 54°

The measure of ∠CDE=54°

Step-by-step explanation:

Given:- Lines CD and DE are tangent to a circle A.

Lines CD and DE are tangents to circle A and intersect at point D.

Arc measures CE = 126°=∠COE

Let o be the centre of circle A

Meet OC and OE which are radius of given circle A.

Now , we know that radius is perpendicular to the tangent.

Therefore, the angle∠OCD=90° and ∠OED=90°

In quadrilateral OCDE

∠OCE+∠OED+∠CDE+∠COE=360°  ( sum of angles of a quadrilateral=360°)

90+90+∠CDE+126=360

∠CDE=360-306

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

d. millimeters per hour.

step-by-step explanation:

we are given that,

carlo used a rain gauge to measure the amount of water his lawn is receiving from the sprinklers.

since, according to the options, we see that,

water cannot be measured by 'miles per hour', 'meters per second' and 'grams per seconds'.

we get that, the water can be measured by 'millimeters per hour'.

hence, the appropriate rate for describing the amount of water is 'millimeters per hour'.