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-Dominant- [34]
3 years ago
5

Which inequality is represented by this graph? ​

Mathematics
1 answer:
vesna_86 [32]3 years ago
7 0

Answer:

number 1

Step-by-step explanation:

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Please help me with this
Troyanec [42]

Answer:

y= \frac{1}{2} +5

4 0
3 years ago
Five times a certain number plus 5 is 45​
ruslelena [56]

Answer:

8

Step-by-step explanation:

8x5=40 + 5 = 45

That's the answer

8 0
3 years ago
-3/4 plus -8/9
Vinil7 [7]

Answer:

-3/4 -8/9

take LCM

<u>-3*9-8*4</u>

    36

<u>-27 -32</u>

  36

-59/36

Step-by-step explanation:

3 0
3 years ago
Suppose a change of coordinates T:R2→R2 from the uv-plane to the xy-plane is given by x=e−2ucos(5v), y=e−2usin(5v). Find the abs
anzhelika [568]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The solution is  \frac{\delta  (x,y)}{\delta (u, v)} | = 10e^{-4u}

Step-by-step explanation:

From the question we are told that

        x =  e^{-2a} cos (5v)

and  y  =  e^{-2a} sin(5v)

Generally the absolute value of the determinant of the Jacobian for this change of coordinates is mathematically evaluated as

     | \frac{\delta  (x,y)}{\delta (u, v)} | =  | \ det \left[\begin{array}{ccc}{\frac{\delta x}{\delta u} }&{\frac{\delta x}{\delta v} }\\\frac{\delta y}{\delta u}&\frac{\delta y}{\delta v}\end{array}\right] |

        = |\ det\ \left[\begin{array}{ccc}{-2e^{-2u} cos(5v)}&{-5e^{-2u} sin(5v)}\\{-2e^{-2u} sin(5v)}&{-2e^{-2u} cos(5v)}\end{array}\right]  |

Let \   a =  -2e^{-2u} cos(5v),  \\ b=-2e^{-2u} sin(5v),\\c =-2e^{-2u} sin(5v),\\d=-2e^{-2u} cos(5v)

So

     \frac{\delta  (x,y)}{\delta (u, v)} | = |det  \left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right] |

=>    \frac{\delta  (x,y)}{\delta (u, v)} | = | a *  b  - c* d |

substituting for a, b, c,d

=>    \frac{\delta  (x,y)}{\delta (u, v)} | =  | -10 (e^{-2u})^2 cos^2 (5v) - 10 e^{-4u} sin^2(5v)|

=>   \frac{\delta  (x,y)}{\delta (u, v)} | =  | -10 e^{-4u} (cos^2 (5v)   + sin^2 (5v))|

=>  \frac{\delta  (x,y)}{\delta (u, v)} | = 10e^{-4u}

7 0
3 years ago
-12x - 12y = 4 3x + 3y = 0
Svetradugi [14.3K]

Answer:

no Solution

Step-by-step explanation:

-12x-12y=4\\ 3x+3y=0

12x-12y=4

add 12y to both sides

12x-12y+12y=4+12y

divid both sides by -12

\frac{-12x}{-12}=\frac{4}{-12}+\frac{12y}{-12}

simplfy

x=-\frac{1+3y}{3}

\mathrm{Substitute\:}x=-\frac{1+3y}{3}

\begin{bmatrix}3\left(-\frac{1+3y}{3}\right)+3y=0\end{bmatrix}

\begin{bmatrix}-1=0\end{bmatrix}

7 0
3 years ago
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