The factored form is (3x-4)(x+9)
Y = mx + b
slope(m) = -2/3
(2,-5)...x = 2 and y = -5
now we sub, we r looking for b, the y int
-5 = -2/3(2) + b
-5 = - 4/3 + b
-5 + 4/3 = b
- 15/3 + 4/3 = b
- 11/3 = b
equation is : y = -2/3x - 11/3...but we need it in standard form
Ax + By = C
y = -2/3x - 11/3
2/3x + y = - 11/3....multiply by 3
2x + 3y = -11 <== standard form
Answer:
The length of the missing side is 7 yards
Step-by-step explanation:
Simply subtract 9 from 16
16-9=7
The length of the missing side is 7 yards
Hope this helps! Plz award Brainliest : )
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
Answer:
I think it’s C
Step-by-step explanation:
