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Nostrana [21]
3 years ago
6

Kira owes Mark $5, and Mark owes Kira $7. Which statement means the same thing?

Mathematics
1 answer:
Mice21 [21]3 years ago
5 0

Answer:

mark owes Kira $2

I think

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What is the answer to <br> –3x2+7=-2x
Damm [24]

Step-by-step explanation:

-6 + 7 = 2x

1 =2x

If you have to find the value of x then

1/2 = x

7 0
3 years ago
Describe the number pattern below<br> 7,345.6; 734.56; 73.456; 5.
KATRIN_1 [288]

Answer:

The place values are decreasing/ moving further to the left one digit at a time. The next three numbers are: 7.3456, 0.73456 and 0.073456.

Step-by-step explanation:

8 0
3 years ago
Please help please ASAP
FrozenT [24]

Answer:

PV = 6 and PT = 3

Hope it will help :) see pic ❤

4 0
3 years ago
Use the method of Lagrange multipliers to find the dimensions of the rectangle of greatest area that can be inscribed in the ell
Tanzania [10]

Answer:

Length (parallel to the x-axis): 2 \sqrt{2};

Height (parallel to the y-axis): 4\sqrt{2}.

Step-by-step explanation:

Let the top-right vertice of this rectangle (x,y). x, y >0. The opposite vertice will be at (-x, -y). The length the rectangle will be 2x while its height will be 2y.

Function that needs to be maximized: f(x, y) = (2x)(2y) = 4xy.

The rectangle is inscribed in the ellipse. As a result, all its vertices shall be on the ellipse. In other words, they should satisfy the equation for the ellipse. Hence that equation will be the equation for the constraint on x and y.

For Lagrange's Multipliers to work, the constraint shall be in the form: g(x, y) =k. In this case

\displaystyle g(x, y) = \frac{x^{2}}{4} + \frac{y^{2}}{16}.

Start by finding the first derivatives of f(x, y) and g(x, y)with respect to x and y, respectively:

  • f_x = y,
  • f_y = x.
  • \displaystyle g_x = \frac{x}{2},
  • \displaystyle g_y = \frac{y}{8}.

This method asks for a non-zero constant, \lambda, to satisfy the equations:

f_x = \lambda g_x, and

f_y = \lambda g_y.

(Note that this method still applies even if there are more than two variables.)

That's two equations for three variables. Don't panic. The constraint itself acts as the third equation of this system:

g(x, y) = k.

\displaystyle \left\{ \begin{aligned} &y = \frac{\lambda x}{2} && (a)\\ &x = \frac{\lambda y}{8} && (b)\\ & \frac{x^{2}}{4} + \frac{y^{2}}{16} = 1 && (c)\end{aligned}\right..

Replace the y in equation (b) with the right-hand side of equation (b).

\displaystyle x = \lambda \frac{\lambda \cdot \dfrac{x}{2}}{8} = \frac{\lambda^{2} x}{16}.

Before dividing both sides by x, make sure whether x = 0.

If x = 0, the area of the rectangle will equal to zero. That's likely not a solution.

If x \neq 0, divide both sides by x, \lambda = \pm 4. Hence by equation (b), y = 2x. Replace the y in equation (c) with this expression to obtain (given that x, y >0) x = \sqrt{2}. Hence y = 2x = 2\sqrt{2}. The length of the rectangle will be 2x = 2\sqrt{2} while the height will be 2y = 4\sqrt{2}. If there's more than one possible solutions, evaluate the function that needs to be maximized at each point. Choose the point that gives the maximum value.

7 0
3 years ago
Which equation are true for x=-2 and x=2 check all that apply
ki77a [65]
1 it is true
{x}^{2} - 4 = 0 \\ {x}^{2} = + 4 \\ x = \sqrt{4} = 2
2 it's not true
{x}^{2} = - 4
3 it's not true
3 {x}^{2} + 12 = 0 \\ 3 {x}^{2} = - 12 \\ {x}^{2} = \frac{ - 12}{3} = - 4 \\
4 it's true
4 {x}^{2} = 16 \\ {x}^{2} = \frac{16}{4} = 4 \\ x = \sqrt{4} = 2
5- it's true
{2(x - 2)}^{2} = 0 \\ {(x - 2)}^{2} = 0 \\ x - 2 = 0 \\ x = 2
4 0
4 years ago
Read 2 more answers
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