The original loan amount is referred to as the_____

Which describes the correct way to convert minutes to hours?

Goshia [24]

Answer:

Divide the number of minutes by 60.

Step-by-step explanation:

7 0

2 years ago

In rectangle PQRS, PR = 18x – 28 and QS = x + 380. Find the value of x and the length of each diagonal.

chubhunter [2.5K]

Answer:

x = 24, PR = 404, QS = 404

Step-by-step explanation:

PR and QS are both diagonals of the rectangle. The diagonals of a rectangle bisect each other and are of equal length. Thus, we can say:

PR = QS

18x - 28 = x + 380

Solving for x, we get:

18x - x = 380 + 28

17x = 408

x = 408/17

x = 24

Thus, PR = 18(24) - 28 = 404 and QS = 24 + 380 = 404

Third choice is right.

7 0

2 years ago

What is the least common denominator of 2/6 and 7/8

Olegator [25]

Answer:

Therefore the least common denominator is 24.

Step-by-step explanation:

2/6 = 8/24

7/8 = 21/24

No more explanation.

6 0

1 year ago

Read 2 more answers

rishi spent 3/4 of an hour each day for 2 days kyle spent 1/4 of an hour each day for 6 days. who spent more time

myrzilka [38]

To find the answer,we need to first find out their hours spent individually.

Rishi spent:
=2×3/4
=6/4hours

Kyle spent
=6×1/4
=6/4hours

As they both spent 6/4 hours,they spent equal time.

Hope it helps!

5 0

2 years ago

Find the critical points of the function f(x, y) = 8y2x − 8yx2 + 9xy. Determine whether they are local minima, local maxima, or

NARA [144]

Answer:

Saddle point: $(0,0)$

Local minimum: $(\frac{3}{8}, -\frac{3}{8})$

Local maxima: $(0,-\frac{9}{8})$ , $(\frac{9}{8},0)$

Step-by-step explanation:

The function is:

$f(x,y) = 8\cdot y^{2}\cdot x -8\cdot y\cdot x^{2} + 9\cdot x \cdot y$

The partial derivatives of the function are included below:

$\frac{\partial f}{\partial x} = 8\cdot y^{2}-16\cdot y\cdot x+9\cdot y$

$\frac{\partial f}{\partial x} = y \cdot (8\cdot y -16\cdot x + 9)$

$\frac{\partial f}{\partial y} = 16\cdot y \cdot x - 8 \cdot x^{2} + 9\cdot x$

$\frac{\partial f}{\partial y} = x \cdot (16\cdot y - 8\cdot x + 9)$

Local minima, local maxima and saddle points are determined by equalizing both partial derivatives to zero.

$y \cdot (8\cdot y -16\cdot x + 9) = 0$

$x \cdot (16\cdot y - 8\cdot x + 9) = 0$

It is quite evident that one point is (0,0). Another point is found by solving the following system of linear equations:

$\left \{ {{-16\cdot x + 8\cdot y=-9} \atop {-8\cdot x + 16\cdot y=-9}} \right.$

The solution of the system is (3/8, -3/8).

Let assume that y = 0, the nonlinear system is reduced to a sole expression:

$x\cdot (-8\cdot x + 9) = 0$

Another solution is (9/8,0).

Now, let consider that x = 0, the nonlinear system is now reduced to this:

$y\cdot (8\cdot y+9) = 0$

Another solution is (0, -9/8).

The next step is to determine whether point is a local maximum, a local minimum or a saddle point. The second derivative test:

$H = \frac{\partial^{2} f}{\partial x^{2}} \cdot \frac{\partial^{2} f}{\partial y^{2}} - \frac{\partial^{2} f}{\partial x \partial y}$

The second derivatives of the function are:

$\frac{\partial^{2} f}{\partial x^{2}} = 0$

$\frac{\partial^{2} f}{\partial y^{2}} = 0$

$\frac{\partial^{2} f}{\partial x \partial y} = 16\cdot y -16\cdot x + 9$

Then, the expression is simplified to this and each point is tested:

$H = -16\cdot y +16\cdot x -9$

S1: (0,0)

$H = -9$ (Saddle Point)

S2: (3/8,-3/8)

$H = 3$ (Local maximum or minimum)

S3: (9/8, 0)

$H = 9$ (Local maximum or minimum)

S4: (0, - 9/8)

$H = 9$ (Local maximum or minimum)

Unfortunately, the second derivative test associated with the function does offer an effective method to distinguish between local maximum and local minimums. A more direct approach is used to make a fair classification:

S2: (3/8,-3/8)

$f(\frac{3}{8} ,-\frac{3}{8} ) = - \frac{27}{64}$ (Local minimum)