Set the equation to 125.
15x+50= 125
Subtract 50.
15x= 75
Divide by 15 into both sides.
x= 5
5 family members went.
I hope this helps!
~cupcake
Answer:
y = 3sin2t/2 - 3cos2t/4t + C/t
Step-by-step explanation:
The differential equation y' + 1/t y = 3 cos(2t) is a first order differential equation in the form y'+p(t)y = q(t) with integrating factor I = e^∫p(t)dt
Comparing the standard form with the given differential equation.
p(t) = 1/t and q(t) = 3cos(2t)
I = e^∫1/tdt
I = e^ln(t)
I = t
The general solution for first a first order DE is expressed as;
y×I = ∫q(t)Idt + C where I is the integrating factor and C is the constant of integration.
yt = ∫t(3cos2t)dt
yt = 3∫t(cos2t)dt ...... 1
Integrating ∫t(cos2t)dt using integration by part.
Let u = t, dv = cos2tdt
du/dt = 1; du = dt
v = ∫(cos2t)dt
v = sin2t/2
∫t(cos2t)dt = t(sin2t/2) + ∫(sin2t)/2dt
= tsin2t/2 - cos2t/4 ..... 2
Substituting equation 2 into 1
yt = 3(tsin2t/2 - cos2t/4) + C
Divide through by t
y = 3sin2t/2 - 3cos2t/4t + C/t
Hence the general solution to the ODE is y = 3sin2t/2 - 3cos2t/4t + C/t
Not sure I think maybe its the second choice. Sorry cant be certain.
Answer:
301.44 mm
Step-by-step explanation:
Diameter of the inside ring of a packing tape = 80 mm
The tape surrounding the ring is 8 mm beyond the ring which means the diameter of the ring including the surrounding will be:
80 + 8 + 8 = 96 mm
and its radius will be = 96/2 = 48.
To find the distance around the outer edge of the tape, we need to calculate the circumference for the ring with the tape on it.
<em>Circumference = 
</em>
Circumference = 2 x 3.14 x 48 = 301.44 mm
Therefore, the distance around the outer edge of the tape is 301.44 mm.
