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3 years ago

A spinner has 3 options and a bag

Mathematics

1 answer:

qaws [65]3 years ago

7 0

Answer:

1/6

Step-by-step explanation:

Spinner = 1 out of 3

Marbles = 2 out of 4

1/3x2/4=2/12=1/6

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PLEASE HELP .. At the beginning of the school year, the number of students in the band was the same as the number of students in

Hunter-Best [27]

I’m thinking 16, but I don’t want to give you the wrong answer..

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4 years ago

Helpanswer question 2

Alona [7]

Answer:

(2,3)

Step-by-step explanation:

-x-(x+1)=-5

-2x-1=-5

-2x=-4

x=2

y=3

5 0

3 years ago

3) z= m - x , for x I need a answer

natulia [17]

Answer:

-z+m=x

You subtract m, then divide z-m by -1.

3 0

4 years ago

Read 2 more answers

X=6. y=5. 12×+3y please help me

IgorLugansk [536]

Answer:

Your answer: 87

Step-by-step explanation:

Since you are given what both variables are in the equation, this problem requires you to plug it in.

12(6)+3(5)= ?

You can either do this in a calculator, or do this one step at a time.

12(6)=72

3(5)=15

72+15=87

7 0

4 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