Ask question

Ask question

All categories

3 years ago

13

The function f(x)=x+n is a polynomial? true or false.

1 answer:

Andrew [12]3 years ago

5 0

False I don’t really know why

You might be interested in

Valerie is going to purchase a new car. The car she wants has a list price of $32,495. Valerie is planning to make a

Anna11 [10]

Answer:

B). $38,821

Step-by-step explanation:

By using several of the known mathematical formulas, and what is given to us at the very bottom (and top) of the answer, we can see that the correct answer is

$38,821

on e2020

4 0

3 years ago

Simplify<br> -3c+5d+3c-d

Rina8888 [55]

Answer:

$4d$

-3x+3c+5d-d

=0+4d

=4d

8 0

3 years ago

Read 2 more answers

Give a counterexample for the statement: All square roots are ration numbers

loris [4]

Counterexamples:

√2, √3, √5, √6, √7, √8, and a great many others, are not rational numbers.

3 0

3 years ago

the u.s.s harry truman is an aircraft carrier that is 1,092 feet long. find the length in ft and in please help

kodGreya [7K]

The U.S.S Harry Truman is 1,092 feet (333 m) long, 257 feet (78 m) wide and is as high as a twenty-four-story building, at 244 feet (74 m)

3 0

3 years ago

Use the variation of parameters method to solve the DR y" + y' - 2y = 1

postnew [5]

Answer:

$y(t)\ =\ C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

Step-by-step explanation:

As given in question, we have to find the solution of differential equation

$y"+y'-2y=1$

by using the variation in parameter method.

From the above equation, the characteristics equation can be given by

$D^2+D-2\ =\ 0$

$=>D=\ \dfrac{-1+\sqrt{1^2+4\times 2\times 1}}{2\times 1}\ or\ \dfrac{-1-\sqrt{1^2+4\times 2\times 1}}{2\times 1}$

$=>\ D=\ -2\ or\ 1$

Since, the roots of characteristics equation are real and distinct, so the complementary function of the differential equation can be by

$y_c(t)\ =\ C_1e^{-2t}+C_2e^t$

Let's assume that

$y_1(t)=e^{-2t}$ $y_2(t)=e^t$

$=>\ y'_1(t)=-2e^{-2t}$ $y'_2(t)=e^t$

and g(t)=1

Now, the Wronskian can be given by

$W=y_1(t).y'_2(t)-y'_1(t).y_2(t)$

$=e^{-2t}.e^t-e^t(-e^{-2t})$

$=e^{-t}+2e^{-t}$

$=3e^{-t}$

Now, the particular solution can be given by

$y_p(t)\ =\ -y_1(t)\int{\dfrac{y_2(t).g(t)}{W}dt}+y_2(t)\int{\dfrac{y_1(t).g(t)}{W}dt}$

$=\ -e^{-2t}\int{\dfrac{e^t.1}{3.e^{-t}}dt}+e^{t}\int{\dfrac{e^{-2t}.1}{3.e^{-t}}dt}$

$=\ -e^{-2t}\int{\dfrac{1}{3}dt}+\dfrac{e^t}{3}\int{e^{-t}dt}$

$=\dfrac{-e^{-2t}}{3}.t-\dfrac{1}{3}$

$=-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

Now, the complete solution of the given differential equation can be given by

$y(t)\ =\ y_c(t)+y_p(t)$

$=C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

5 0

3 years ago