D) c=3p+2 3 caterpillars per plant and 2 on ground
Answer:
k = ln (6/5)
Step-by-step explanation:
for
f(x)=A*exp(kx)+B
since f(0)=1, f(1)=2
f(0)= A*exp(k*0)+B = A+B = 1
f(1) = A*exp(k*1)+B = A*e^k + B = 2
assuming k>0 , the horizontal asymptote H of f(x) is
H= limit f(x) , when x→ (-∞)
when x→ (-∞) , limit f(x) = limit (A*exp(kx)+B) = A* limit [exp(kx)]+B* limit = A*0 + B = B
since
H= B = (-4)
then
A+B = 1 → A=1-B = 1 -(-4) = 5
then
A*e^k + B = 2
5*e^k + (-4) = 2
k = ln (6/5) ,
then our assumption is right and k = ln (6/5)
A) x – 7 = -7(6y + 5)
x – 7 = -42y + -35
x – 7 + 35 = -42y
x + 28 = -42y
y = (1/42)x – 14/21
c) y – 7 = 7(x + 5)
y – 7 = 7x + 35
y = 7x + 42
b) y – 7 = -7(x + 5)
y – 7 = -7x -35
y = -7x – 28
d) y + 7 = 7(x + 5)
y + 7 = 7x + 35
y = 7x + 28
Answer:
=2f^10
Step-by-step explanation:
We have that
<span>f(x) = (3/2)*x + 4
case 1) for x=-8</span>f(-8) = (3/2)*(-8) + 4--------> -8
point (-8,-8)
case 2) for x=-4f(-4) = (3/2)*(-4) + 4--------> -2
point (-4,-2)
case 3) for x=-2f(-2) = (3/2)*(-2) + 4--------> 1
point (-2,1)
case 4) for x=0f(0) = (3/2)*(0) + 4--------> 4
point (0,4)
case 5) for x=2f(2) = (3/2)*(2) + 4--------> 7
point (2,7)
case 6) for x=4f(4) = (3/2)*(4) + 4--------> 10
point (4,10)
using a graph tool
see the attached figure