To encourage john to work harder in math his mother said she would pay him 10
cents for each right answer and subtract 5 cents for each wrong answer. if he earned
20 cents after doing 32 problems, how many problems did john get right? how
many did he get wrong? how many would he have to get right to earn more than a
dollar?

he got 12 questions right

he got 20 questions wrong.

Step-by-step explanation:

Let x represent the number of questions that he got right.

Let y represent the number of questions that he got wrong.

His mother said she would pay him 10 cents for each right answer. This means that if he gets x questions right, he would earn $10x.

His mother said she would subtract 5 cents for each wrong answer. This means that if he gets y questions wrong, he would earn

- $5y.

If he solved 32 problems, it means that

x + y = 32

If he earned 20 cents after doing 32 problems,it means that

10x - 5y = 20 - - - - - - - - -1

Substituting x = 32 - y into equation 1, it becomes

10(32 - y) - 5y = 20

320 - 10y - 5y = 20

- 10y - 5y = 20 - 320

- 15y = - 300

y = - 300/ -15 = 20

Substituting y = 20 into x = 32 - y, it becomes.

x = 32 - 20 = 12

To get more than a dollar, he needs to earn at least 110 cents. He already has 20 cents. He needs to get 9 right to earn over a dollar.

If he gets 9 right, he would earn 90 cents more. 90 + 20 = 110 cents = $1.1

the parachutist hits the ground at t= 211.43 sec

Step-by-step explanation:

from Newton's second law:

Force = m*g - b*v = m*dv/dt

∫dt = ∫dv/(g - b/m*v) = - (m/b) ln (g - b/m*v) +C

t = - (m/b) ln (g - b/m*v) +C

assuming that he starts at rest (when he jumps) for t =0 → v=0

0 = - (m/b) ln (g - b/m*0) +C → C=(m/b) ln g

thus

t = (m/b) ln [g/(g - b/m*v)]

e^(b/m*t) = g/(g - b/m*v)

e^-(b/m*t) = 1- b/(m*g)*v

v= m*g/b [1-e^-(b/m*t) ]

then

dh/dt=v

∫dh = ∫v * dt = v=  ∫ m*g/b [1-e^-(b/m*t) ] dt =  m*g/b*[t + m/b *e^-(b/m*t)]

h = m*g/b*[t + m/b *e^-(b/m*t)]

then

a) before it opens b₁= 15N sec/m , thus

h = 75*9.8/15*[ 60 + 75/15 *e^-(15/75*60)] = 2940 m

b) after if opens  b₂= 105N sec/m , thus for the remaining meters

4000 - 2940 = 75*9.8/105*[ t + 75/105 *e^-(105/75*t)] =

since the equation is not linear we can neglect the exponencial term ( assume that reaches fast the terminal velocity since b is high), then

4000 - 2940 = 75*9.8/105* t

t= 151.43 sec

then to prove our assumption

additional distance neglected = 75/105 *e^-(105/75*151.43 )] = 2,36*10⁻³⁷ m ≈ 0

then our assumption is right

thus total time is

t total =  60 sec + 151.43 sec = 211.43 sec

therefore the parachutist hits the ground at t= 211.43 sec