Evaluaté p/4+pq when p=8 and q=6

Mathematics

2 answers:

dusya [7]3 years ago

4 0

So the right answer is 50.

Look at the attached picture

Hope it will help you....

Good luck on your assignment

~pragya

Alexxandr [17]3 years ago

4 0

Answer: 50

Step-by-step explanation:

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Emil is itemizing deductions on his federal income tax return and had $1900 in non-reimbursed work expenses last year. If his AG

a_sh-v [17]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

Below are the choices that can be found from other source:

A.$340
B.$1560 
C.$38 
D.$1862

Below is the solution:

78000 x .0.02=1560
1900 - 1560=340

5 0

3 years ago

Read 2 more answers

PLEASE HELP ASAP!!!!

Setler [38]

Piecewise Function is like multiple functions with a speific/given domain in one set, or three in one for easier understanding, perhaps.

To evaluate the function, we have to check which value to evalue and which domain is fit or perfect for the three functions.

Since we want to evaluate x = -8 and x = 4. That means x^2 cannot be used because the given domain is less than -8 and 4. For the cube root of x, the domain is given from -8 to 1. That meand we can substitute x = -8 in the cube root function because the cube root contains -8 in domain but can't substitute x = 4 in since it doesn't contain 4 in domain.

Last is the constant function where x ≥ 1. We can substitute x = 4 because it is contained in domain.

Therefore:

$\large{ \begin{cases} f( - 8 ) = \sqrt[3]{ - 8} \\ f(4) = 3 \end{cases}}$

The nth root of a can contain negative number only if n is an odd number.

$\large{ \begin{cases} f( - 8 ) = \sqrt[3]{ - 2 \times - 2 \times - 2} \\ f(4) = 3 \end{cases}} \\ \large{ \begin{cases} f( - 8 ) = - 2\\ f(4) = 3 \end{cases}}$