Find the coordinates of n if n is exactly halfway between m (-10,-13) and d (-7,-7)

Answers

Answer 1

Answer: I think it’s Midpoint is (xM, yM) = (-8.5, -10) but I might be wrong but this is what I got.

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Translate into a variable expression.the square of the difference between a number n and fifty

If the distance from C to C’ is 8 and the scale factor is 5. What is the distance from A to C? A. 8/5
B. 5/8
C. 10
D. 2

Answers

Answer : B

Explanation:

Find the exact values of the numbers c that satisfy the conclusion of the Mean Value Theorem for the interval [−2, 2]. (Enter your answers as a comma-separated list.)

Answers

Answer:

The answer is " $\bold{c= \pm (2)/(√(3))}$ "

Step-by-step explanation:

If the function is:

$\to f'(x) = 3x^2-2 \n\n\to f'(c) = 3c^2-2$

points are:

$\to -2 \leq x \leq2$

use the mean value theorem:

$\to f'(c) = ( f(b)- f(a))/(b-a)$

$= ( f(2)- f(-2))/(2-(-2))\n\n= (4-(-4) )/(4)\n\n= (8)/(4)\n\n= 2$

$\to 3c^2-2=2 \n\n\to 3c^2=4 \n\n\to c^2=(4)/(3) \n\nc= \pm (2)/(√(3))$

Final answer:

The Mean Value Theorem states that for a continuous and differentiable function on a closed interval, there exists at least one 'c' within that interval where the average change rate equals the instantaneous rate at 'c'. In the given case of interval [-2,2], to find 'c', first calculate the average slope between the points (f(2)-f(-2))/4. Then equate this average slope to the derivative 'f'(c). The solution(s) to this equation are the c values for this problem.

Explanation:

The subject of this question pertains to the Mean Value Theorem in Calculus. According to this theorem, if a function f is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one number c in the open interval (a, b) such that the average rate of change over the interval equals the instantaneous rate of change at c.

In the given case, we're trying to find the 'c' value for the interval [-2,2]. First, we need to find the average slope between the two points. Assuming f is your function, that would be (f(2)-f(-2))/ (2 - -2). Subtract the function values of the two points and divide by the total interval length. Next, we need to see where this average slope equals the instantaneous slope 'f'(c), this entails solving the equation 'f'(c) = (f(2)-f(-2))/4. The solution to this equation will be the c values that satisfy the Mean value theorem within the provided interval.

Learn more about Mean Value Theorem here:

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A plant produces 500 units/hour of an item with dimensions of 4” x 6” x 2”. The manager wants to store two weeks of supply in containers that measure 3 ft x 4 ft x 2 ft. (Note: She can store the units in the containers such as that the 4” dimension aligns with either the 3 ft width or the 4 ft length of the box, whichever allows more units to be stored.) A minimum of 2 inches of space is required between adjacent units in each direction. If the containers must be stacked 4-high, and the warehouse ceiling is 9 feet above the floor, then determine the amount of floor space required just for storage.

Answers

Answer:

564 ft²

Step-by-step explanation:

To account for the extra space between units, we can add 2" to every unit dimension and every box dimension to figure the number of units per box.

Doing that, we find the storage box dimensions (for calculating contents) to be ...

3 ft 2 in × 4 ft 2 in × 2 ft 2 in = 38 in × 50 in × 26 in

and the unit dimensions to be ...

(4+2)" = 6" × (6+2)" = 8" × (2+2)" = 4"

A spreadsheet can help with the arithmetic to figure how many units will fit in the box in the different ways they can be arranged. (See attached)

When we say the "packing" is "462", we mean the 4" (first) dimension of the unit is aligned with the 3' (first) dimension of the storage box; the 6" (second) dimension of the unit is aligned with the 4' (second) dimension of the storage box; and the 2" (third) dimension of the unit is aligned with the 2' (third) dimension of the storage box. The "packing" numbers identify the unit dimensions, and their order identifies the corresponding dimension of the storage box.

We can see that three of the four allowed packings result in 216 units being stored in a storage box.

If storage boxes are stacked 4 deep in a 9' space, the 2' dimension must be the vertical dimension, and the floor area of each stack of 4 boxes is 3' × 4' = 12 ft². There are 216×4 = 864 units stored in each 12 ft² area.

If we assume that 2 weeks of production are 80 hours of production, then we need to store 80×500 = 40,000 units. At 864 units per 12 ft² of floor space, we need ceiling(40,000/864) = 47 spaces on the floor for storage boxes. That is ...

47 × 12 ft² = 564 ft²

of warehouse floor space required for storage.

_____

The second attachment shows the top view and side view of units packed in a storage box.

Determine the maximized area of a rectangle that has a perimeter equal to 56m by creating and solving a quadratic equation. What is the length and width?