Answer:
82 cm
Step-by-step explanation:
In rectangles diagonals are equal and bisect each other
AO = BO
5x + 1 = 4x + 9
Subtract 1 from both sides
5x = 4x + 9 -1
5x = 4x + 8
Subtract 4x from both the sides
5x - 4x = 8
x = 8
AO = 5x + 1
= 5*8 +1
= 40 + 1
AO= 41 cm
Diagonal = 2*41 = 82 cm
Answer:
Yes this is a linear function f(x) = ex+30
e = hourly rate
x=hours
f(x) = total earnings
You would take random numbers preferable smaller ones such as -1,0,1,2
You would put those in the place of x
Then you would square the value of x. After that you subtract 4. Whatever you come out with is your y value. Then you put the x value on the x axis and from that point count up or down to the number you need on the y axis. Together that would be your point
Answer:
Please find attached a drawing of the triangles ΔRST and EFG showing the angles
The angle on ΔEFG that would prove the triangles are similar is ∠F = 25°
Step-by-step explanation:
In order to prove that two triangles are similar, two known angles of each the triangles need to be shown to be equal
Given that triangle ∠R and ∠S of triangle ΔRST are 95° and 25°, respectively, and that ∠E of ΔEFG is given as 90°, then the corresponding angle on ΔEFG to angle ∠S = 25° which is ∠F should also be 25°
Therefore, the angle on ΔEFG that would prove the triangles are similar is ∠F = 25°.
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
