All categories

2 years ago

Solve the following equation for p: 6/p=x+a

1 answer:

6 0

$\dfrac 6p = x+a\\\\\implies \dfrac 6p \times\dfrac 16 = (x+a) \times\dfrac 16 ~~~~~~;\left[\text{Multiply both sides by}~ \dfrac 16 \right]\\\\\implies \dfrac 1p = \dfrac{x+a}6\\\\\implies p= \dfrac{6}{x+a}~~~~~~~~~~~~~~~~~~~~;[\text{Cross multiply}]$

You might be interested in

A group of 4 friends likes to golf together, and each friend keeps track of her all-time lowest score in a single round. Their l

ale4655 [162]

To find the median you list out all the numbers and find the middle.

80, 90, 100

With angie getting 75, it goes from "80, 90, 100" to "75, 80, 90, 100"

This changes the median from 90 to 85.

Mean is the average of all the numbers. The average of the first set of numbers (Found by adding all of the numbers up and dividing by the amount of numbers present) is 90.

The mean of the second set is 86.25

The answer would be A, both the mean and median will decrease.

5 0

3 years ago

Read 2 more answers

My 4 time posting this can anyone please help me please 14 and 16

Katena32 [7]

Answer:

14. 780

16. 274

hopes this helps have a great night

Step-by-step explanation:

5 0

3 years ago

|1/2b-8|=|1/4b-1|<br> b=____ and ____

Deffense [45]

Answer:

b = 12 and 28

Step-by-step explanation:

The absolute value equation |1/2b-8| = |1/4b-1| resolves to a piecewise linear function with three pieces. There are two solutions.

<h3>Domains</h3>

The absolute value function on the left has a turning point where its value is zero:

1/2b -8 = 0

b -16 = 0

b = 16

The absolute value function on the right has a turning point where its value is zero:

1/4b -1 = 0

b -4 = 0

b = 4

For b > 16, both absolute value functions are identity functions. In this domain, the equation is ...

1/2b -8 = 1/4b -1

For 4 < b < 16, the function on the left negates its argument, so the equation in this domain is ...

-(1/2b -8) = 1/4b -1

For b < 4, both functions negate their arguments, so the equation in this domain is ...

-(1/2b -8) = -(1/4b -1)

For the purpose of finding the value of b, this is effectively identical to the equation for b > 16. (The value of b does not change if we multiply both sides of the equation by -1.)

<h3>Solutions</h3>

<u>Domain b < 4 ∪ 16 < b</u>

1/2b -8 = 1/4b -1

2b -32 = b -4 . . . . . . . . multiply by 4

b = 28 . . . . . . . . . . . . add 32-b to both sides

This solution is in the domain of the equation, so is one of the solutions to the original equation.

<u>Domain 4 < b < 16</u>

-(1/2b -8) = 1/4b -1 . . . . equation in this domain

-2b +32 = b -4 . . . . . . multiply by 4

36 = 3b . . . . . . . . . . . add 2b+4 to both sides

12 = b . . . . . . . . . . . . divide by 3

This solution is in the domain of the equation, so is the other solution to the original equation.

<h3>Graph</h3>

For the purposes of the graph, we have define the function g(b) to be the difference of the two absolute value functions. The solutions are found where g(x) = 0, the x-intercepts. The graph shows those to be ...

b = 12 and b = 28

<em>Additional comment</em>

Defining g(b) = |1/2b-8| -|1/4b-1|, we can rewrite it as ...

$g(b)=\begin{cases}7-\dfrac{1}{4}b&\text{for }b < 4\\-\dfrac{3}{4}b+9&\text{for }4\le b < 16\\\dfrac{1}{4}b-7&\text{for }16\le b\end{cases}$