J = 14 + P
J+2 = 3 (P+2) Sub in J = 14 + P
14 + P + 2 = 3P + 6
16+P = 3P + 6
10 = 2P
P = 5 then J = 14+P = 19
(75 + 97 + 60 + x) / 4 = 80
and
(75 + 97 + 60 + x) /4 = 89 Solve for'x' in both equations to find the range of scores....
The are of this square is (2+x) * (2+x) = 7
4 + 4x + x ^2 = 7 Now subtract 4 from both sides.........and.......
FOUR of the common difference ( from term 5 to 9) is 11 to 7 so the common d is -1
9th is 7 16th will be 7* -1 more = 0
Use tan 70 = opp/adj
tan 70 = height/12
y = x^5-1 /3 find the inverse function
x = y^5-1 /3
(3x +1)1/5 =y this is f-1
(3(-31/96)+1 )1/5 =y
y = 1/2
This situation forms a right triangle......the line of sight to the nest is the hypotenuse and one leg os 10 feet while the other leg is the height of the nest.
tan 55o = opp/adj
tan 55 = height/10 ft
1.428 = height / 10 solve for height.
c x ( 1+i)n -1
________ x (1+i) = 922.80 n = 9 i = .06/12 = .005 c = 100
i
https://web2.0calc.com/questions/pls-help-i-am-stuck-on-this-really-badly
mid point is -4, 6
slope is (8-4) / (-3 - -5) = 4/2 = 2 perpindicualr slope -1/2
y = mx + b
6 = -1/2 (-4) + b b = 4
y = -1/2 x + 4