Ask question

Ask question

All categories

1 year ago

15

What are the parts of an algebraic expression, and how do they relate to polynomial expressions?

1 answer:

Arlecino [84]1 year ago

3 0

The parts of algebraic expressions related to polynomials are variables and coefficients.

<h3>What are the parts of algebraic expressions?</h3>

The parts of algebraic expressions are;

Variables which are letters that represent numbers
Coefficients are numbers that multiply the variables
Constant is a number that is not multiplied by any variable

Polynomials are algebraic expressions composed of variables, constants and exponents, that are combined using the mathematical operations such as addition, subtraction, and multiplication

Polynomials consists of variables and coefficients. The variables in polynomials are also called indeterminates. The coefficients also multiply this variables.

Thus, the parts of algebraic expressions related to polynomials are variables and coefficients.

Learn more about algebraic expressions here:

brainly.com/question/4344214

#SPJ1

You might be interested in

What would an equation be for 6 bags of flyers each bag containing 12 flyers

enyata [817]

Answer:

6x12

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

I NEED HELP ASAP!<br> WILL MARK BRAINLIEST!!

Burka [1]

I don't see a mistake. You solve in parentheses first, then you solve exponents. If the exponent is 0 in this context, the answer would be 0.

5 0

2 years ago

Read 2 more answers

The total monthly profit for a firm is P(x)=6400x−18x^2− (1/3)x^3−40000 dollars, where x is the number of units sold. A maximum

wlad13 [49]

Answer:

Maximum profits are earned when x = 64 that is when 64 units are sold.

Maximum Profit = P(64) = 2,08,490.666667$

Step-by-step explanation:

We are given the following information: $P(x) = 6400x - 18x^2 - \frac{x^3}{3} - 40000$ , where P(x) is the profit function.

We will use double derivative test to find maximum profit.

Differentiating P(x) with respect to x and equating to zero, we get,

$\displaystyle\frac{d(P(x))}{dx} = 6400 - 36x - x^2$

Equating it to zero we get,

$x^2 + 36x - 6400 = 0$

We use the quadratic formula to find the values of x:

$x = \displaystyle\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$ , where a, b and c are coefficients of $x^2, x^1 , x^0$ respectively.

Putting these value we get x = -100, 64

Now, again differentiating

$\displaystyle\frac{d^2(P(x))}{dx^2} = -36 - 2x$

At x = 64, $\displaystyle\frac{d^2(P(x))}{dx^2} < 0$

Hence, maxima occurs at x = 64.

Therefore, maximum profits are earned when x = 64 that is when 64 units are sold.

Maximum Profit = P(64) = 2,08,490.666667$

6 0

2 years ago

If the lines of a system of equations never intersect then, then the system has no____

eduard

Answer:

solution

Step-by-step explanation:

7 0

3 years ago

How to change 0.225 into a fraction

Aloiza [94]

I believe it’s 9/30

8 0

2 years ago