Which equation correctly shows the relationship between the numbers 6,560 and 656?

Find the missing side in the similar figure below

xeze [42]

Answer:

30

Step-by-step explanation:

20 divided by 12 is 5/3.

18x5/3 is 30, meaning the missing side is 30.

7 0

3 years ago

The daily cost of hiring a plumber , y , to work x hours on a repair project can be modeled using a linear function. The plumber

Paladinen [302]

Answer:

The range of the function is $80\leq y\leq 440$

Step-by-step explanation:

Consider the provided information.

The plumber charges a fixed cost of $80 plus an additional cost of $45 per hour.

Let plumber works for x hours, then the required linear function will be:

$y=45x+80$

The plumber works a maximum of 8 hours per day.

That means the minimum value of x is 0 and maximum value of x is 8.

We need to find the range of the function.

Range of the function is the set of y values.

Substitute x=0 in above function,

$y=45(0)+80$

$y=80$

Now substitute x=8 in above function.

$y=45(8)+80$

$y=360+80$

$y=440$

Hence, the range of the function is $80\leq y\leq 440$

3 0

3 years ago

A graphing calculator is recommended. A function is given. g(x) = x4 − 5x3 − 14x2 (a) Find all the local maximum and minimum val

Taya2010 [7]

Answer:

The local maximum and minimum values are:

Local maximum

$g(0) = 0$

Local minima

$g(5.118) = -350.90$

$g(-1.368) = -9.90$

Step-by-step explanation:

Let be $g(x) = x^{4}-5\cdot x^{3}-14\cdot x^{2}$ . The determination of maxima and minima is done by using the First and Second Derivatives of the Function (First and Second Derivative Tests). First, the function can be rewritten algebraically as follows:

$g(x) = x^{2}\cdot (x^{2}-5\cdot x -14)$

Then, first and second derivatives of the function are, respectively:

First derivative

$g'(x) = 2\cdot x \cdot (x^{2}-5\cdot x -14) + x^{2}\cdot (2\cdot x -5)$

$g'(x) = 2\cdot x^{3}-10\cdot x^{2}-28\cdot x +2\cdot x^{3}-5\cdot x^{2}$

$g'(x) = 4\cdot x^{3}-15\cdot x^{2}-28\cdot x$

$g'(x) = x\cdot (4\cdot x^{2}-15\cdot x -28)$

Second derivative

$g''(x) = 12\cdot x^{2}-30\cdot x -28$

Now, let equalize the first derivative to solve and solve the resulting equation:

$x\cdot (4\cdot x^{2}-15\cdot x -28) = 0$

The second-order polynomial is now transform into a product of binomials with the help of factorization methods or by General Quadratic Formula. That is:

$x\cdot (x-5.118)\cdot (x+1.368) = 0$