They will hold 112.25 cm^3 more. Find the volume of the second can with the new side lengths and subtract the orginal volume from the new one.
Answer:
BC = 1.71
Step-by-step explanation:
well to start we have to know the relationship between angles, legs and the hypotenuse in a right triangle
α = 70°
a: adjacent = BC
h: hypotenuse = 5
sin α = o/h
cos α= a/h
tan α = o/a
we see that it has (angle, adjacent, hypotenuse)
we look at which meets those data between the sine, cosine and tangent
is the cosine
cos α = a/h
Now we replace the values and solve
cos 70 = a/5
0.34202 = a/5
0.34202 * 5 = a
1.7101 = a
round to the neares hundredth
a = 1.7101 = 1.71
BC = 1.71
Step-by-step explanation:
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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
