All categories

3 years ago

Determine which expressions are polynomials. Explain your reasoning.

Mathematics

1 answer:

Tom [10]3 years ago

8 0

Answer:

A, B

Step-by-step explanation:

A polynomial is an expression containing two or more algebraic terms. They are often the sum of several terms containing different powers (exponents) of variables and never contains square roots or division by a variable.

In your case, option A contains two terms that with second and first power of x. Therefore, the expression in A is a polynomial. Option B contains three terms that with second, first power of x and coefficient. This is a polynomial too. Option C represents the absolute value of x and doesn't represent a polynomial.

You might be interested in

Help on this math question ‍♀️

noname [10]

Answer:

the 2nd 3rd and 4th answers are the correct ones

Step-by-step explanation:

hope this helps you

4 0

2 years ago

How much interest would you have if you deposited $200 that if you deposited $150?

Andrew [12]

Answer:

Difference = 206 - 154.5 = $51.5

In short, Your Answer would be: $51.5

7 0

2 years ago

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'

vodka [1.7K]

Answer:

a) $v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

b) $0$

c) $a(t) = x''(t) = v'(t) =6t-12$

When the acceleration is 0 we have:

$6t-12=0, t =2$

And if we replace t=2 in the velocity function we got:

$v(t) = 3(2)^2 -12(2) +9=-3$

Step-by-step explanation:

For this case we have defined the following function for the position of the particle:

$x(t) = t^3 -6t^2 +9t -5 , 0\leq t\leq 10$

Part a

From definition we know that the velocity is the first derivate of the position respect to time and the accelerations is the second derivate of the position respect the time so we have this:

$v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

Part b

For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:

$v(t) = 3t^2 -12t +9$

We can factorize this function as $v(t)= 3 (t^2- 4t +3)=3(t-3)(t-1)$

So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is $0\leq t\leq 10$ we need to analyze the following intervals: