Given:
Angle A is 4 degrees greater than the measure of Angle B. Both angles are complementary
Complementary angles have a sum of 90°
Angle A = x + 4° ; Angle B = x
x + 4° + x = 90°
2x = 90° - 4°
2x = 86°
x = 86° ÷ 2
x = 43° ANGLE B.
Angle A = x + 4° ⇒ 43° + 4 = 47°
Given:
Angle D is 5 times the measure of Angle E. These angles are supplementary. This means that their sum is 180°
Angle D = 5x ; Angle E = x
5x + x = 180°
6x = 180°
x = 180° ÷ 6
x = 30° Angle E.
Angle D = 5x = 5(30) = 150°
Using Laplace transform we have:L(x')+7L(x) = 5L(cos(2t))sL(x)-x(0) + 7L(x) = 5s/(s^2+4)(s+7)L(x)- 4 = 5s/(s^2+4)(s+7)L(x) = (5s - 4s^2 -16)/(s^2+4)
=> L(x) = -(4s^2 - 5s +16)/(s^2+4)(s+7)
now the boring part, using partial fractions we separate 1/(s^2+4)(s+7) that is:(7-s)/[53(s^2+4)] + 1/53(s+7). So:
L(x)= (1/53)[(-28s^2+4s^3-4s^2+35s-5s^2+5s)/(s^2+4) + (-4s^2+5s-16)/(s+7)]L(x)= (1/53)[(4s^3 -37s^2 +40s)/(s^2+4) + (-4s^2+5s-16)/(s+7)]
denoting T:= L^(-1)and x= (4/53) T(s^3/(s^2+4)) - (37/53)T(s^2/(s^2+4)) +(40/53) T(s^2+4)-(4/53) T(s^2/s+7) +(5/53)T(s/s+7) - (16/53) T(1/s+7)
56348372828282828348484959382829594
Answer:
oh! thanks for points
but this is yr question or what????
Answer:
Step-by-step explanation:
<u>Property of regular polygons:</u>
- Sum of Interior Angles = (n−2) × 180°, where n is number of sides
<u>In our case we have:</u>
- 158.8×n = (n - 2)×180
- 180n - 158.8n = 360
- 21.2n = 360
- n = 360/21.2
- n = 17
