5. x + y = 2(2k) + 1

the value of y varies directly with x, and when x = 1/4 , y = 25. What is the value of y when x is 2 1/3 need help ASAP

Grace [21]

Answer:

$y=233\frac{1}{3}$

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $y/x=k$ or $y=kx$

In this problem we have

For $x=1/4, y=25$

Find the value of k (constant of proportionality)

$y/x=k$ -----> $k=25/(1/4)=100$

The linear equation is

$y=100x$

Find the value of y when the value of $x=2\frac{1}{3}$

substitute the value of x in the equation but first convert to an improper fraction

$x=2\frac{1}{3}=\frac{2*3+1}{3}=\frac{7}{3}$

$y=100(\frac{7}{3})=\frac{700}{3}$

convert the result in mixed number

$\frac{700}{3}=\frac{699}{3}+\frac{1}{3}=233\frac{1}{3}$

3 0

4 years ago

If a, b, 72, (a+b) are consecutive terms of an A.P., find the values of a and b.

mel-nik [20]

Answer:

a = 36, b = 54

Step-by-step explanation:

Since the terms are in arithmetic progression then there is a common difference d between consecutive terms , that is

b - a = 72 - b ( add b to both sides )

2b - a = 72 → (1)

and

a + b - 72 = b - a ( subtract b - a from both sides )

2a - 72 = 0 ( add 72 to both sides )

2a = 72 ( divide both sides by 2 )

a = 36

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Substitute a = 36 into (1) and evaluate for b

2b - 36 = 72 ( add 36 to both sides )

2b = 108 ( divide both sides by 2 )

b = 54

-----------------

Thus a = 36 and b = 54

The sequence is therefore 36, 54, 72, 90 , .....

5 0

3 years ago

Read 2 more answers

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.