$e\cdot e^x -e^{-2}=-2$
$\implies e^{x+1}=e^{-2}-2$
note that RHS is negative. (because e with negative exponent is less than 1)
and LHS is always positive.
so there cannot be any solution
The formula that is useful for solving both of these problems is ...
![(v_2)^2-(v_1)^2=2ad\qquad\text{a=acceleration, d=distance}\\\\ \text{where $v_1$ and $v_2$ are initial and final velocities}](https://tex.z-dn.net/?f=%28v_2%29%5E2-%28v_1%29%5E2%3D2ad%5Cqquad%5Ctext%7Ba%3Dacceleration%2C%20d%3Ddistance%7D%5C%5C%5C%5C%20%5Ctext%7Bwhere%20%24v_1%24%20and%20%24v_2%24%20are%20initial%20and%20final%20velocities%7D)
9. Given v₁=15, a=9.8, d=10, find v₂.
... (v₂)² = 15² + 2·9.8·10
... v₂ = √421 ≈ 20.5 . . . . m/s
10. Given d=12 m when a=-9.8 m/s² and v₂=0, find d when a=0.17·(-9.8 m/s²).
The formula tells us that d=(v₁)²/(2a), which is to say that the distance is inversely proportional to the acceleration. If acceleration is 0.17 times that on earth, distance will be 1/0.17 ≈ 5.88 times that on earth.
(12 m)/0.17 ≈ 70.6 m
Answer:
-21 - 15g
Step-by-step explanation:
You just have to distribute the -3 into the expression.
-3(7 + 5g) = -21 -15g
⇒
C......................................
Plug -6 in for x and solve.
y = (-1/3)(-6) + 5
y = 2 + 5
y = 7