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nlexa [21]
3 years ago
5

The latitude value of Toronto is 43.70 degrees, and the longitude value is 79.40 degrees. The latitude value of Melbourne is -37

.81 degrees, and the longitude value is 144.96 degrees. The two cities are degrees apart in latitude. The two cities are degrees apart in longitude.
Mathematics
1 answer:
SVETLANKA909090 [29]3 years ago
5 0
I don't really know what is it
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Laura creates a rectangular prism with wooden cubes. Each cube has an edge length of 3/4 inch. she uses a total of 240 cubes. th
oksano4ka [1.4K]
2/4 would be the fraction
3 0
3 years ago
Show that f(x) f(y) - f(x+y), where f(x)= [cos x - sin x o] [sin x cos x o] [o o 1]
Naddik [55]

Answer:

The answer is below

Step-by-step explanation:

Show that f(x) f(y) = f(x+y)

From trigonometric:

sin(x + y) = sinxcosy + cosxsiny

sin(x - y) = sinxcosy - cosxsiny

cos(x + y) = cosxcosy - sinxsiny

cos(x - y) = cosxcosy + sinxsiny

f(x)=\left[\begin{array}{ccc}cosx&-sinx&0\\sinx&cosx&0\\0&0&1\end{array}\right] ,f(y)=\left[\begin{array}{ccc}cosy&-siny&0\\siny&cosy&0\\0&0&1\end{array}\right] \\\\\\f(x)f(y)=\left[\begin{array}{ccc}cosxcosy-sinxsiny&-cosxsiny-sinxcosy&0\\sinxcosy+cosxsiny&-sinxsiny+cosxcosy&0\\0&0&1\end{array}\right] \\\\\\f(x)f(y)=\left[\begin{array}{ccc}cos(x+y)&-sin(x+y)&0\\sin(x+y)&cos(x+y)&0\\0&0&1\end{array}\right] \\\\\\

f(x+y)=\left[\begin{array}{ccc}cos(x+y)&-sin(x+y)&0\\sin(x+y)&cos(x+y)&0\\0&0&1\end{array}\right] \\\\\\Therefore\ f(x)f(y)=f(x+y)

4 0
3 years ago
A) Use the definition of Laplace transform to find L{f }. (Do the integrals.) For what values of s is L{f } defined?f(t) = (2t+1
kiruha [24]

For the given function f(t) = (2t + 1) using definition of Laplace transform the required solution is L(f(t))s = [ ( 2/s²) + ( 1/s) ].

As given in the question,

Given function is equal to :

f(t) = 2t + 1

Simplify the given function using definition of Laplace transform we have,

L(f(t))s = \int\limits^\infty_0 {f(t)e^{-st} } \, dt

          =  \int\limits^\infty_0[2t +1] e^{-st} dt

          = 2\int\limits^\infty_0 te^{-st} + \int\limits^\infty_0e^{-st} dt

         = 2 L(t) + L(1)

L(1) = \int\limits^\infty_0e^{-st} dt

     = (-1/s) ( 0 -1 )

     = 1/s , ( s >  0)

2L ( t ) = 2\int\limits^\infty_0 te^{-st}

        =  2[t\int\limits^\infty_0 e^{-st} - \int\limits^\infty_0 ({(d/dt)(t) \int\limits^\infty_0e^{-st} \, dt )dt]

        =  2/ s²

Now ,

L(f(t))s = 2 L(t) + L(1)

          = 2/ s² + 1/s

Therefore, the solution of the given function using Laplace transform the required solution is L(f(t))s = [ ( 2/s²) + ( 1/s) ].

Learn more about Laplace transform here

brainly.com/question/14487937

#SPJ4

8 0
1 year ago
two years of local internet service cost $685 including the installation fee of $85 what is the monthly fee?
zimovet [89]
2 years is the same as 24 months, so the equation would be:
685 = 24x + 85, where x is the monthly fee.
600 = 24x
600/24 = x
x = 25
The monthly fee is $25.
7 0
3 years ago
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