Answer:
<A = 44
Step-by-step explanation:
A triangle is 180 degrees
(x+59) + (x+51) + 84 = 180 >> add everything on the left side
2x + 194 = 180 >> subtract both sides by 194
2x = -14 >> divide both sides by 2 to get x alone
x = -7
since you are finding A<, substitute
<A = x + 51
<A = (-7) + 51
<A = 44
Answer:
a. 
b. 
Step-by-step explanation:
The initial value problem is given as:

Applying laplace transformation on the expression 
to get ![L[{y+y'} ]= L[{7 + \delta (t-3)}]](https://tex.z-dn.net/?f=L%5B%7By%2By%27%7D%20%5D%3D%20L%5B%7B7%20%2B%20%5Cdelta%20%28t-3%29%7D%5D)

Taking inverse of Laplace transformation
![y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]](https://tex.z-dn.net/?f=y%28t%29%20%3D%207%20L%5E%7B-1%7D%20%5B%20%5Cdfrac%7B1%7D%7B%28s%2B1%29%7D%5D%20%2B%20L%5E%7B-1%7D%20%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D%20%5C%5C%20%5C%5C%20y%28t%29%20%3D%207L%5E%7B-1%7D%20%5B%5Cdfrac%7B%28s%2B1%29-s%7D%7Bs%28s%2B1%29%7D%5D%20%2BL%5E%7B-1%7D%20%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D%20%5C%5C%20%5C%5C%20y%28t%29%20%3D%207L%5E%7B-1%7D%20%5B%5Cdfrac%7B1%7D%7Bs%7D-%5Cdfrac%7B1%7D%7Bs%2B1%7D%5D%20%2B%20L%5E%7B-1%7D%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D%20%5C%5C%20%5C%5C%20y%28t%29%20%3D%207%20%5B1-e%5E%7B-t%7D%20%5D%20%2B%20L%5E%7B-1%7D%20%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D)
![L^{-1}[\dfrac{e^{-3s}}{s+1}]](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D)
![L^{-1}[\dfrac{1}{s+1}] = e^{-t} = f(t) \ then \ by \ second \ shifting \ theorem;](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5B%5Cdfrac%7B1%7D%7Bs%2B1%7D%5D%20%3D%20e%5E%7B-t%7D%20%20%3D%20f%28t%29%20%5C%20then%20%5C%20by%20%5C%20second%20%5C%20shifting%20%5C%20theorem%3B)
![L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D%20%3D%20%5Cleft%20%5C%7B%20%7B%7Bf%28t-3%29%20%5C%20%5C%20%5C%20t%3E3%7D%20%5Catop%20%7B0%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%5C%20%5C%20%20%5C%20t%20%3C3%7D%7D%20%5C%20%5C%20%5C%20%20%5Cright.)
![L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D%20%3D%20%5Cleft%20%5C%7B%20%7B%7Be%5E%7B%28-t-3%29%7D%20%5C%20%5C%20%5C%20t%3E3%7D%20%5Catop%20%7B0%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%5C%20%5C%20%20%5C%20t%20%3C3%7D%7D%20%5C%20%5C%20%5C%20%20%5Cright.)

= 
Recall that:
![y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]](https://tex.z-dn.net/?f=y%28t%29%20%3D%207%20%5B1-e%5E%7B-t%7D%20%5D%20%2B%20L%5E%7B-1%7D%20%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D)
Then



slope = - 
the equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y-intercept )
rearrange 2x + 3y = - 12 into this form
subtract 2x from both sides
3y = - 2x - 12 ( divide all terms by 3 )
y = -
x - 4 ← in slope- intercept form
with slope m = - 
1) The function is
3(x + 2)³ - 32) The
end behaviour is the
limits when x approaches +/- infinity.3) Since the polynomial is of
odd degree you can predict that
the ends head off in opposite direction. The limits confirm that.
4) The limit when x approaches negative infinity is negative infinity, then
the left end of the function heads off downward (toward - ∞).
5) The limit when x approaches positive infinity is positivie infinity, then
the right end of the function heads off upward (toward + ∞).
6) To graph the function it is important to determine:
- x-intercepts
- y-intercepts
- critical points: local maxima, local minima, and inflection points.
7)
x-intercepts ⇒ y = 0⇒ <span>
3(x + 2)³ - 3 = 0 ⇒ (x + 2)³ - 1 = 0
</span>
<span>⇒ (x + 2)³ = -1 ⇒ x + 2 = 1 ⇒
x = - 1</span>
8)
y-intercepts ⇒ x = 0y = <span>3(x + 2)³ - 3 =
3(0 + 2)³ - 3 = 0 - 3×8 - 3 = 24 - 3 =
21</span><span>
</span><span>
</span><span>9)
Critical points ⇒ first derivative = 0</span><span>
</span><span>
</span><span>i) dy / dx = 9(x + 2)² = 0
</span><span>
</span><span>
</span><span>⇒ x + 2 = 0 ⇒
x = - 2</span><span>
</span><span>
</span><span>ii)
second derivative: to determine where x = - 2 is a local maximum, a local minimum, or an inflection point.
</span><span>
</span><span>
</span><span>
y'' = 18 (x + 2); x = - 2 ⇒ y'' = 0 ⇒ inflection point.</span><span>
</span><span>
</span><span>Then the function does not have local minimum nor maximum, but an
inflection point at x = -2.</span><span>
</span><span>
</span><span>Using all that information you can
graph the function, and I
attache the figure with the graph.
</span>