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Amiraneli [1.4K]
3 years ago
5

PLEASE HELP!!!!

Mathematics
2 answers:
Katena32 [7]3 years ago
6 0

Answer:

C

Step-by-step explanation:

Tatiana [17]3 years ago
6 0

Answer:

Pretty sure it's

D. None of the above

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Which classification best describes the following system of equations? 3x+6y-12z=36, x+2y-4z=12, 4x+8-16z=48
Andrews [41]
(1) 3x+6y-12z=36
(2) x+2y-4z=12
(3) 4x+8y-16z=48

The first equation (1) is the second equation (2) multiplied by 3:
(2) x+2y-4z=12→3(x+2y-4z=12)→3x+6y-12z=36 (1)

The third equation (3) is the second equation (2) multiplied by 4:
(2) x+2y-4z=12→4(x+2y-4z=12)→4x+8y-16z=48 (3)

The equations are linearly dependent, Then the system of equations is dependent, and then consistent too.
5 0
4 years ago
Math homework geometry please help
Ghella [55]

Answer:

A

Step-by-step explanation:

The theorem we use for this is

a(a + b) = c(c + d)

Solve for d.

a(a + b) = c { }^{2}  + cd

a(a + b) -  {c}^{2}

\frac{a(a + b) -  {c}^{2} }{c}

A is the answer

3 0
3 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
Really don’t understand guys help please
aleksandr82 [10.1K]
Looks like it’s equal to
5 0
4 years ago
What is X of 4x+6 = 8x+3
Karolina [17]

Answer:

3/4

Step-by-step explanation:

4x+6 = 8x+3

Subtract 4x from both sides to get x on the same side

6 = 4x+3

Subtract 3 from both sides:

3 = 4x

Divide by 4:

x = 3/4

5 0
3 years ago
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