Can someone please help with these problems I’m having trouble pls

When the chords intersect inside the circle the sum of the vertical angles and the sum of the referenced arcs are the same.  Since the vertical angles are congruent, each is half the sum of the referenced arcs.

Similarly when the chords intersect outside the circle, the angle is half the difference of the arc measures.

It's the same with a tangent; in that case there's just not a part of the arc that's not counted on the side with the tangent.

OK, that's how.  Let's try some.

1.  x = (88 + 86)/2 = 87

2. x = (150 - 60)/2 = 45

3. The angle is exactly half the subtended arc,

x = 110/2 = 55

4. The referenced arc is 360 - 220 = 140 degrees and x is half, 70.

5.  Outside, half the difference, x = (90-20)/2 = 35

6.  Angle outside, inner arc 6 degrees, vertical angle is 90 degrees.

90 = (x - 6)/2

180 = x - 6

x = 186

7.  There's a special rule for this case but we can treat it as outside, little arc x, big arc 360-x,

60 = (360 - x  - x)/2

120 = 360 - 2x

2x = 240

x = 120

8. Inner arc x, outer arc is 360-140-x = 220-x.  Half the difference is 38,

38 = (220 - x - x)/2

76 = 220 - 2x

2x = 220 - 76

x = (220 - 76)/2 = 72

9.  x is formed by a tangent and a radius, so a right angle, x=90

y is subtended by a 35 degree angle so y=70

10.  Remaining triangle angle is 180 - 62 - 28 = 90.  Arc that subtends it is double, 180.  That means the long side is a diameter.

11.  Remaining triangle angle is 180 - 50 - 70 = 60. x=120, y=100, z=140

12.  Remaining arc 360 - 170 - 90 = 100.  Angles are half.

x=45, y=50, z=85

.24

step-by-step explanation:

the probability is   .24

the first player had the highest probability of getting a silver marble

the game could be made fair by adding back the marble that was drawn after each draw

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

=====================================

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).