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victus00 [196]
3 years ago
6

Classify each polynomial by its degree and number of terms

Mathematics
2 answers:
Soloha48 [4]3 years ago
6 0
You have that:

 1) 7x/4+3 (Name using degree: First degree polynomial),(Name using number of terms:binomial)

 2) 5.2x^2-4x+2.5 (Name using degree: Quadratic polynomial),(Name using number of terms:trinomial)

 3) 3/5 (Name using degree: zero degree polynomial),(Name using number of terms:monomial)

 4) 0.75x^2 (Name using degree: Quadratic degree polynomial),(Name using number of terms:monomial).
r-ruslan [8.4K]3 years ago
3 0

Answer with explanation:

We know that a polynomial is classified <u>based on the degree</u> as follows:

Constant-

If it has a degree zero.

and the equation of such a polynomial is:

   p(x)=a

Linear--

It has degree 1.

and the equation of such a polynomial is:   p(x)=ax+b

Quadratic--

If it has degree two and the equation of the polynomial is given by:

     p(x)=ax^2+bx+c

<u>Based on number of Terms--</u>

Monomial--

If it has just one term.

Binomial--

If it has two terms.

Trinomial--

If it has three terms.

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Round $56.29 to the nearest dollar.
jeka57 [31]

Answer:

$56

Step-by-step explanation:

5 0
3 years ago
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Integrate t sec^2 (2t) dt
Neporo4naja [7]
F = t ⇨ df = dt 
dg = sec² 2t dt ⇨ g = (1/2) tan 2t 
⇔ 
integral of t sec² 2t dt = (1/2) t tan 2t - (1/2) integral of tan 2t dt 
u = 2t ⇨ du = 2 dt 

As integral of tan u = - ln (cos (u)), you get : 

integral of t sec² 2t dt = (1/4) ln (cos (u)) + (1/2) t tan 2t + constant 
integral of t sec² 2t dt = (1/2) t tan 2t + (1/4) ln (cos (2t)) + constant 
integral of t sec² 2t dt = (1/4) (2t tan 2t + ln (cos (2t))) + constant ⇦ answer 
8 0
3 years ago
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Consider functions of the form f(x)=a^x for various values of a. In particular, choose a sequence of values of a that converges
sleet_krkn [62]

Answer:

A. As "a"⇒e, the function f(x)=aˣ tends to be its derivative.

Step-by-step explanation:

A. To show the stretched relation between the fact that "a"⇒e and the derivatives of the function, let´s differentiate f(x) without a value for "a" (leaving it as a constant):

f(x)=a^{x}\\ f'(x)=a^xln(a)

The process will help us to understand what is happening, at first we rewrite the function:

f(x)=a^x\\ f(x)=e^{ln(a^x)}\\ f(x)=e^{xln(a)}\\

And then, we use the chain rule to differentiate:

f'(x)=e^{xln(a)}ln(a)\\ f'(x)=a^xln(a)

Notice the only difference between f(x) and its derivative is the new factor ln(a). But we know  that ln(e)=1, this tell us that as "a"⇒e, ln(a)⇒1 (because ln(x) is a continuous function in (0,∞) ) and as a consequence f'(x)⇒f(x).

In the graph that is attached it´s shown that the functions follows this inequality (the segmented lines are the derivatives):

if a<e<b, then aˣln(a) < aˣ < eˣ < bˣ < bˣln(b)  (and below we explain why this happen)

Considering that ln(a) is a growing function and ln(e)=1, we have:

if a<e<b, then ln(a)< 1 <ln(b)

if a<e, then aˣln(a)<aˣ

if e<b, then bˣ<bˣln(b)

And because eˣ is defined to be the same as its derivative, the cases above results in the following

if a<e<b, then aˣ < eˣ < bˣ (because this function is also a growing function as "a" and "b" gets closer to e)

if a<e, then aˣln(a)<aˣ<eˣ ( f'(x)<f(x) )

if e<b, then eˣ<bˣ<bˣln(b) ( f(x)<f'(x) )

but as "a"⇒e, the difference between f(x) and f'(x) begin to decrease until it gets zero (when a=e)

3 0
3 years ago
Mr. Fuller wants to put fencing around his rectangular-shaped yard. the width of the yard is 55 feet and the length is 75 feet.
AlekseyPX
260 feet. 55+55=110, 75+75=150, 150+110=260
5 0
1 year ago
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Given the graph below, which of the following statements is true?
nirvana33 [79]

Answer:

D) The graph does not represent a one-to-one function because the y-values between 0 to 2 are paired with multiple x-values.

Step-by-step explanation:

Let's find the ordered pairs.

(-4, 4), (-3, 2), (-2, 0), (-1, 0.5), (0, 1), (1, 1.5), (2, 2), (3, 0)

The graph passed through above points.

In the x-coordinates -3, 2 gives the same output. Therefore, the given function is not one-to-one.

Therefore, the answer D)

The graph does not represent a one-to-one function because the y-values between 0 to 2 are paired with multiple x-values.

Hope you will understand the concept.

Thank you.

7 0
2 years ago
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