Answer:
5 or -5
Step-by-step explanation:
3+2=5
-1(3+2)=-5
Answer:
3x^2 or 3x squared
Step-by-step explanation:
4x^2-x^2=3x^2
4x squared-x squared equals x squared
when subtracting squared it's like subtracting normal numbers
Answer:
D is not a function.
Step-by-step explanation:
The first option is a function because it makes up a straight line. The second option is also a function because it is also known as y = -3/2x + 2 creating a slope function. The third option is also a function because there is no more than one point for each x value. The last option is not a function because the number 2 is going for 2 variables A and B and you can't have them with more than one so D is your answer.
Answer:
The correct answer is 15 cm.
Step-by-step explanation:
Let the width of the required poster be a cm.
We need to have a 6 cm margin at the top and a 4 cm margin at the bottom. Thus total margin combining top and bottom is 10 cm.
Similarly total margin combining both the sides is (4+4=) 8 cm.
So the required printing area of the poster is given by {( a-10 ) × ( a - 8) } 
This area is equal to 125 as per as the given problem.
∴ (a - 10) × (a - 8) = 125
⇒ 
- 18 a +80 -125 =0
⇒ - 18 a -45 = 0
⇒ (a-15) (a-3) = 0
By law of trichotomy the possible values of a are 15 and 3.
But a=3 is absurd as a 
4.
Thus the required answer is 15 cm.
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
