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3 years ago

What binomial do you have to add to the polynomial x^2+y^2–2xy+1 to get a polynomial: not containing the variable y?

Mathematics

1 answer:

zvonat [6]3 years ago

4 0

Since you do not want the variable y, you must eliminate $y^{2}-2xy$

Therefore, to eliminate it, you must have the opposite of it, making the answer $2xy-y^{2}$

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On a trip, you had to change your money from dollars to British pounds. You got 560560 pounds for 800800 dollars. How many pound

Vlad [161]

First you need to divide 800,800 by 300,300 because you need to know how these two amounts are related, in order to know how the amounts of pounds you are converting are related.
800,800÷300,300= 2.66666.... (recurring)
If $300,300 goes into $800,800 2.66666 times, then the amount of pounds you want to find goes into £560,560 2.66666 times as well. Let's write this as an equation to solve for x (where x is the amount of pounds you will obtain for $560,560):
2.66666x = 560,560
Divide both sides by 2.66666
x = 210,210
Therefore, you will get £210,210 for $300,000

7 0

3 years ago

Help please! I’ll give you brainlist

11111nata11111 [884]

Answer:

1,007 calories in total

Step-by-step explanation:

8 0

3 years ago

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

3 0

3 years ago

2(5-3v)+9v=28 step by step please

Irina-Kira [14]

First, distribute the 2 across the parenthesis.
10-3v+9v=28
Combine like terms on the left.
10+6v=28
Start to get the numbers on the right and the v alone on the left by moving the 10 to the right.
6v=28-10
6v=18
Divide to get the v alone.
v=3

Hope that helps.

4 0

3 years ago

In what base is the sum of 8 and 5 equal to 14?

LuckyWell [14K]

Nothing. It would have to be 5 and 8 to be 13... Not 14 and it would be math. :)

4 0

4 years ago