600 milimeters by 500 millimeters divided by 25 centimeters

1 answer:

8 0

As 600 mm and 500 mm convert to 60 cm and 50 cm respectively, our problem here becomes (60 cm)(50 cm) / (25 cm), or 120 cm

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Determine the value for c on {2, 5} that satisfies the Mean Value Theorem for f(x) = x^2 - 3 / x -1

yanalaym [24]

Answer:

B. 3.

Step-by-step explanation:

OK lets try again.

The slope of the secant = slope of the tangent at a certain point ( The Mean Value Theorem).

Slope of the secant = f(5) - f(2) / (5 - 2)

= [(25-3) / (5-1) - (4-3) / (2-1)] / 3

= (22/4 - 1) / 3

= 9/2 / 3

= 9/6

= 3/2.

The derivative at c = the slope of the tangent at c.

Finding the derivative:

f'(x) = [2x(x - 1) - (x^2 - 3) ]/ (x - 1)^2 (where x = c).

= (x^2 - 2x + 3)/ (x - 1)^2 = the slope.

So equating the slopes:

(x^2 - 2x + 3) / (x - 1)^2 = 3/2

2x^2 - 4x + 6 = 3x^2 - 6x + 3

x^2 - 2x - 3 = 0

(x - 3)(x + 1) = 90

x = 3 , -1

x can't be -1 because we are working between the values 2 and 5 so

x = c = 3.

4 0

3 years ago

The graph of a quadratic function having the form f(x) = ax? + bx + c passes

insens350 [35]

Answer:

f(-3) = -20

Step-by-step explanation:

We observe that the given x-values are 3 units apart, and that the x-value we're concerned with is also 3 units from the first of those given. So, a simple way to work this is to consider the sequence for x = 6, 3, 0, -3. The corresponding sequence of f(x) values is ...

34, 10, -8, ?

The first differences of these numbers are ...

10 -34 = -24

-8 -10 = -18

And the second difference is ...

-18 -(-24) = 6

For a quadratic function, second differences are constant. This means the next first-difference will be ...

? -(-8) = -18 +6

? = -12 -8 = -20

The value of the function at x=-3 is -20.

_____

The attachment shows using a graphing calculator to do a quadratic regression on the given points. The graph can then be used to find the point of interest. There are algebraic ways to do this, too, but they are somewhat more complicated than the 5 addition/subtraction operations we needed to find the solution. (Had the required x-value been different, we might have chosen a different approach.)