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3 years ago

8x-5y=21 simultaneous linear equation2x+9y=-5

Mathematics

1 answer:

olya-2409 [2.1K]3 years ago

8 0

Answer:

$x=2$ and $y=-1$

Step-by-step explanation:

If two equations are given, we can solve one of the equation in terms of single variable and put it in the other equation which will further give the value of x and y.

For the given equations:

$8x-5y=21$ Equation:1

$2x+9y=-5$ Equation:2

Solving equation multiply equation 2 with 4 on both the sides

$8x+36y=-20$ Equation 3

Subtracting Equation:3 from Equation:1

$8x-5y-(8x+36y)=21-(-20)\\8x-5y-8x-36y=21+20\\-41y=41\\y=-1$

Putting value of 'y' in Equation:2 which will give the value of x

$2x+9y=-5\\2x+9(-1)=-5\\2x-9=-5\\2x=4\\x=2$

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2 years ago

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3 years ago

Mrs. Howard is hosting a catered dinner party to celebrate her daughter's graduation. The catering company she plans to use char

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8 0

2 years ago

Read 2 more answers

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

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

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2 years ago

PLEASE HELP MATH

Novosadov [1.4K]

I believe C is your answer.
I hope this helps!

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2 years ago

Read 2 more answers