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faltersainse [42]
3 years ago
10

Which ordered pairs in the form  (x, y) are solutions to the equation 7x−5y=28 ?Solation 7x−5y=28 ? A. (−6, −14) B. (7,9) C. (4,

10) D.(-1,-7)
Mathematics
1 answer:
Mkey [24]3 years ago
7 0
Probe every pair:

pair                 x          y           7x - 5y

A. (-6,-14)      -6       -14          7(-6) - 5(-14) = -42 + 70 = 28 => it is a solution

B. (7,9)           7          9          7(7) - 5(9) = 49 - 45 = 4 => not a solution

C. (4,10)        4          10        7(4) - 5(10) = 28 - 50 = - 22 => not a solution

C. (-1,-7)        -1          -7       7(-1) - 5(-7) = -7 + 35 = 28 => it is a solution

Answer: option A. (-6,-14) and option D. (-1,-7)
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angle U and angle W are vertical angles. If measurement angle u = 6x+11 and measurement angle W = 10x-9, find measurement angle
IgorC [24]

The measure of angle U will be equal to the measure of angle W.

Because W is pronounced "double u", W = 2 * U.

Therefore U = W and U = 2 * W, so both U and W = 0.


Just kidding.

The correct answer is the smallest prime number greater than 40.

8 0
3 years ago
Let the graph of f(x) represent the cost in thousands of dollars to feed the zoo animals daily, where x is the number of animals
pickupchik [31]

Answer:

<u>There are 200 animals, and the cost is $8,000 daily</u>

Step-by-step explanation:

y=f(x) represents the cost in thousands

x is the number of animals measured in hundreds

so the point (2,8) means

x = 2

y = f(2) = 8

so it means  for 2*100=200 animals, the cost is 8*1000=8 000

The correct answer is then

There are 200 animals, and the cost is $8,000 daily

hope this helps

5 0
3 years ago
Find the critical points of the surface f(x, y) = x3 - 6xy + y3 and determine their nature.​
Vedmedyk [2.9K]

Compute the gradient of f.

\nabla f(x,y) = \left\langle 3x^2 - 6y, -6x + 3y^2\right\rangle

Set this equal to the zero vector and solve for the critical points.

3x^2-6y = 0 \implies x^2 = 2y

-6x+3y^2=0 \implies y^2 = 2x \implies y = \pm\sqrt{2x}

\implies x^2 = \pm2\sqrt{2x}

\implies x^4 = 8x

\implies x^4 - 8x = 0

\implies x (x-2) (x^2 + 2x + 4) = 0

\implies x = 0 \text{ or } x-2 = 0 \text{ or } x^2 + 2x + 4 = 0

\implies x = 0 \text{ or } x = 2 \text{ or } (x+1)^2 + 3 = 0

The last case has no real solution, so we can ignore it.

Now,

x=0 \implies 0^2 = 2y \implies y=0

x=2 \implies 2^2 = 2y \implies y=2

so we have two critical points (0, 0) and (2, 2).

Compute the Hessian matrix (i.e. Jacobian of the gradient).

H(x,y) = \begin{bmatrix} 6x & -6 \\ -6 & 6y \end{bmatrix}

Check the sign of the determinant of the Hessian at each of the critical points.

\det H(0,0) = \begin{vmatrix} 0 & -6 \\ -6 & 0 \end{vmatrix} = -36 < 0

which indicates a saddle point at (0, 0);

\det H(2,2) = \begin{vmatrix} 12 & -6 \\ -6 & 12 \end{vmatrix} = 108 > 0

We also have f_{xx}(2,2) = 12 > 0, which together indicate a local minimum at (2, 2).

3 0
2 years ago
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