What is the value of x if |x-1)=x-2 ?

Heeellllpppppp

jolli1 [7]

Answer:

y=-50x+-10

Step-by-step explanation:

I'm assuming you need this answer in the y=mx+b format.

3 0

3 years ago

the school newspaper has a circulation of 500 and sells for $.035 a copy. the students decide to raise the price to increase the

Alik [6]

Let x be the number of times they raise the price on the newspaper. Then the new cost of the newspaper is

$.35+.05x$

Let y be the newspaper they sell, then the income will be

$y(.35+.05x)$

Now, we know that the circulation is of 500, assuming that they sold every newspaper at the original price now the number the will sell will be

$y=500-20x$

Plugging the value of y in the first expression we have that the income will be

$\begin{gathered} f(x)=(500-20x)(.35+\text{0}.5x)=175+25x-7x-x^2 \\ =-x^2+18x+175 \end{gathered}$

Then the income is given by the function

$f(x)=-x^2+18x+175$

To find the maximum value of this functions (thus the maximum income) we need to take the derivative of the function,

$\begin{gathered} \frac{df}{dx}=\frac{d}{dx}(-x^2+18x+175) \\ =-2x+18 \end{gathered}$

no we equate the derivative to zero and solve for x.

$\begin{gathered} \frac{df}{dx}=0 \\ -2x+18=0 \\ -2x=-18 \\ x=\frac{-18}{-2} \\ x=9 \end{gathered}$

This means that we have an extreme value of the function when x=9. Now we need to find out if this value is a maximum or a minimum. To do this we need to take the second derivative of the function, then

$\begin{gathered} \frac{d^2f}{dx^2}=\frac{d}{dx}(\frac{df}{dx}) \\ =\frac{d}{dx}(-2x+18) \\ =-2 \end{gathered}$

Since the second derivative is negative in the point x=9, we conclude that this value is a maximum of the function.

With this we conclude that the number of times that they should raise the price to maximize the income is 9. This means that they will raise the price of the newspaper (9)($0.05)=$0.45.

Therefore the price to maximize the income is $0.35+$0.45=$0.80.

4 0

1 year ago

Find the solution of 2/3x+1/4=-4/3

wolverine [178]

These questions can be daunting at first, but they're pretty simple to solve.

First, we need to establish a common denominator. We have 2 / 3, 1 / 4, and
-4 / 3. The least common denominator we can get is by multiplying 4 and 3 together to get 12. So we will change the denominator as follows;

2 / 3, 1 / 4, -4 / 3 = 8 / 12, 3 / 12, -16 / 12

Now we can put these back into the equation.

8/12x + 3/12 = -16/12
8x + 3 = -16

It's simple math from here on out, but I'll show the process. What we can basically do now is take away the denominator because it doesn't matter now that it's common.

Subtract 3 from both sides. Now we have 8x = -19

Dividing by 8 on both sides of the equation will get you your answer.

x = -19/8

Hope this helps!

6 0

3 years ago

Which list of ordered pairs is NOT a function? Mark only one oval. {(1, 1), (2, 3), (3, 4), (4,5)} ){(1, 1), (2, 2), (3, 3), (4,

Mrac [35]

Answer:

{(1, 1), (1, 3), (2, 3), (3,4)} is not a function.

Step-by-step explanation:

The lists are:

a) {(1, 1), (2, 3), (3, 4), (4,5)}

b) {(1, 1), (2, 2), (3, 3), (4,4)}

c) {(1, 1), (3, 1), (4,2), (5, 3) }

d) {(1, 1), (1, 3), (2, 3), (3,4)}

Firs, let's define what is a function.

A function is a relationship that maps elements from a set (the domain, is the set of the inputs) into elements of another set (the range, the set of the outputs) such that each element of the domain can be mapped into only one element of the range.

Then the list where we have the same input mapped into two different outputs, will not be a function.

Where in a ordered pair (x, y)

x is the input, and y is the output.

In this case, the list that is not a function is:

d) {(1, 1), (1, 3), (2, 3), (3,4)}

because we have the input x = 1, mapped into two different outputs.

5 0

3 years ago

Part 1. Using the two functions listed below, insert numbers in place of the letters a, b, c, and d so that f(x) and g(x) are in

dem82 [27]

There is no b unfortunately

Find inverse of x+a

x+a=y

interchAnge

y+a=x
y=x-a

Inverse is

f^{-1}x=x-a

Now

x-a=cx-d
x-cx=a-d
x(1-c)=a-d
x=a-d/1-c

So

You can have any number according to this but

c must not be 1
a and d must not be equal .
All three should belongs to integers set

Note:-

As there is no b given in question (May be missed in typing) Part-2 's equation can't be executed and part 3 also

5 0

3 years ago