X=-2→y=7(6)^(-2+2)+1=7(6)^0+1=7(1)+1=7+1→y=8; (x,y)=(-2,8)
x=-5→y=7(6)^(-5+2)+1=7(6)^(-3)+1=7/6^3+1=7/216+1=0.0324+1→y=1.0324→(x,y)=(-5,1.0324)
Answer: Graph 3
Answer:
5hrs
Step-by-step explanation:
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Answer:
373.8mmHg
Step-by-step explanation:
a =height (in km) above sea level,
the pressure P(a) (in mmHg) is approximated given as
P(a) = 760e–0.13a .
To determine the atmospheric pressure at 5.458 km, then we will input into the equation
P(5.458km) = 760e–0.13a .
= 760e^(-0.13×5.458)
=760e^-(0.70954)
= 760×0.4919
=373.8mmHg
Therefore, the atmospheric pressure at 5.458 km is 373.8mmHg
Answer:
y = 2x − 1
Step-by-step explanation:
By eliminating the parameter, first solve for t:
x = 4 + ln(t)
x − 4 = ln(t)
e^(x − 4) = t
Substitute:
y = t² + 6
y = (e^(x − 4))² + 6
y = e^(2x − 8) + 6
Taking derivative using chain rule:
dy/dx = e^(2x − 8) (2)
dy/dx = 2 e^(2x − 8)
Evaluating at x = 4:
dy/dx = 2 e^(8 − 8)
dy/dx = 2
Writing equation of line using point-slope form:
y − 7 = 2 (x − 4)
y = 2x − 1
Now, without eliminating the parameter, take derivative with respect to t:
x = 4 + ln(t)
dx/dt = 1/t
y = t² + 6
dy/dt = 2t
Finding dy/dx:
dy/dx = (dy/dt) / (dx/dt)
dy/dx = (2t) / (1/t)
dy/dx = 2t²
At the point (4, 7), t = 1. Evaluating the derivative:
dy/dx = 2(1)²
dy/dx = 2
Writing equation of line using point-slope form:
y − 7 = 2 (x − 4)
y = 2x − 1