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Amanda [17]
3 years ago
9

What is the first four common multiples for 5 and 8

Mathematics
2 answers:
marta [7]3 years ago
7 0

Step-by-step explanation:

The first four common multiples of 5 are...

5, 10, 15, 20, 25, 30, 30, 40, 45, 50

The first four common multiples of 8 are...

8, 16, 24, 32, 40, 48, 56, 64, 72, 80

ololo11 [35]3 years ago
3 0

Answer: 5, 10, 15, 20, 25

8, 16, 24, 32

Step-by-step explanation:

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ACCORDING TO THE MODEL

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4 years ago
Verify that the conclusion of Clairaut’s Theorem holds, that is, uxy = uyx, u=tan(2x+3y)
choli [55]

Answer: Hello mate!

Clairaut’s Theorem says that if you have a function f(x,y) that have defined and continuous second partial derivates in (ai, bj) ∈ A

for all the elements in A, the, for all the elements on A you get:

\frac{d^{2}f }{dxdy}(ai,bj) = \frac{d^{2}f }{dydx}(ai,bj)

This says that is the same taking first a partial derivate with respect to x and then a partial derivate with respect to y, that taking first the partial derivate with respect to y and after that the one with respect to x.

Now our function is u(x,y) = tan (2x + 3y), and want to verify the theorem for this, so lets see the partial derivates of u. For the derivates you could use tables, for example, using that:

\frac{d(tan(x))}{dx} = 1/cos(x)^{2} = sec(x)^{2}

\frac{du}{dx}  =  \frac{2}{cos^{2}(2x + 3y)} = 2sec(2x + 3y)^{2}

and now lets derivate this with respect to y.

using that \frac{d(sec(x))}{dx}= sec(x)*tan(x)

\frac{du}{dxdy} = \frac{d(2*sec(2x + 3y)^{2} )}{dy}  = 2*2sec(2x + 3y)*sec(2x + 3y)*tan(2x + 3y)*3 = 12sec(2x + 3y)^{2}tan(2x + 3y)

Now if we first derivate by y, we get:

\frac{du}{dy}  =  \frac{3}{cos^{2}(2x + 3y)} = 3sec(2x + 3y)^{2}

and now we derivate by x:

\frac{du}{dydx} = \frac{d(3*sec(2x + 3y)^{2} )}{dy}  = 3*2sec(2x + 3y)*sec(2x + 3y)*tan(2x + 3y)*2 = 12sec(2x + 3y)^{2}tan(2x + 3y)

the mixed partial derivates are equal :)

7 0
3 years ago
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