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3 years ago

All about simulitious equations

Mathematics

1 answer:

Korolek [52]3 years ago

5 0

Answer:

On occasions you will come across two or more unknown quantities, and two or more equations

relating them. These are called simultaneous equations and when asked to solve them you

must find values of the unknowns which satisfy all the given equations at the same time.

Step-by-step explanation:

1. The solution of a pair of simultaneous equations

The solution of the pair of simultaneous equations

3x + 2y = 36, and 5x + 4y = 64

is x = 8 and y = 6. This is easily verified by substituting these values into the left-hand sides

to obtain the values on the right. So x = 8, y = 6 satisfy the simultaneous equations.

2. Solving a pair of simultaneous equations

There are many ways of solving simultaneous equations. Perhaps the simplest way is elimination. This is a process which involves removing or eliminating one of the unknowns to leave a

single equation which involves the other unknown. The method is best illustrated by example.

Example

Solve the simultaneous equations 3x + 2y = 36 (1)

5x + 4y = 64 (2) .

Solution

Notice that if we multiply both sides of the first equation by 2 we obtain an equivalent equation

6x + 4y = 72 (3)

Now, if equation (2) is subtracted from equation (3) the terms involving y will be eliminated:

6x + 4y = 72 − (3)

5x + 4y = 64 (2)

x + 0y = 8

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1. Approximate the given quantity using a Taylor polynomial with n3.

Jet001 [13]

Answer:

See the explanation for the answer.

Step-by-step explanation:

Given function:

$f(x) = x^{1/4}$

The n-th order Taylor polynomial for function f with its center at a is:

$p_{n}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(n)}a}{n!} (x-a)^{n}$

As n = 3 So,

$p_{3}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(3)}a}{3!} (x-a)^{3}$

$p_{3}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(3)}a}{6} (x-a)^{3}$

$p_{3}(x) = a^{1/4} + \frac{1}{4a^{ 3/4} } (x-a)+ (\frac{1}{2})(-\frac{3}{16a^{7/4} } ) (x-a)^{2} + (\frac{1}{6})(\frac{21}{64a^{11/4} } ) (x-a)^{3}$

$p_{3}(x) = 81^{1/4} + \frac{1}{4(81)^{ 3/4} } (x-81)+ (\frac{1}{2})(-\frac{3}{16(81)^{7/4} } ) (x-81)^{2} + (\frac{1}{6})(\frac{21}{64(81)^{11/4} } ) (x-81)^{3}$

$p_{3} (x)$ = 3 + 0.0092592593 (x - 81) + 1/2 ( - 0.000085733882) (x - 81)² + 1/6

(0.0000018522752) (x-81)³

$p_{3} (x)$ = 0.0092592593 x - 0.000042866941 (x - 81)² + 0.00000030871254

(x-81)³ + 2.25

Hence approximation at given quantity i.e.

x = 94

Putting x = 94

$p_{3} (94)$ = 0.0092592593 (94) - 0.000042866941 (94 - 81)² +

0.00000030871254 (94-81)³ + 2.25

= 0.87037 03742 - 0.000042866941 (13)² + 0.00000030871254(13)³ +

2.25

= 0.87037 03742 - 0.000042866941 (169) +

0.00000030871254(2197) + 2.25

= 0.87037 03742 - 0.007244513029 + 0.0006782414503 + 2.25

$p_{3} (94)$ = 3.113804102621

Compute the absolute error in the approximation assuming the exact value is given by a calculator.

Compute $\sqrt[4]{94}$ as $94^{1/4}$ using calculator

Exact value:

$E_{a}$ (94) = 3.113737258478

Compute absolute error:

Err = | 3.113804102621 - 3.113737258478 |

Err (94) = 0.000066844143

If you round off the values then you get error as:

|3.11380 - 3.113737| = 0.000063

Err (94) = 0.000063

If you round off the values up to 4 decimal places then you get error as:

|3.1138 - 3.1137| = 0.0001

Err (94) = 0.0001

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3 years ago

Find decimal notation 7/20

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Answer:

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3 years ago

Read 2 more answers

The solutions to a certain quadratic equation are x= -4 and X = 3. Write the equation in standard form below.

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Answer:

x² + x - 12 = 0

Step-by-step explanation:

If x = -4 and x = 3 are solutions to a quadratic equation, then

(x + 4)(x - 3) = 0

Expand the brackets: x² - 3x + 4x - 12 = 0

Combine like terms: x² + x - 12 = 0

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2 years ago

All about simulitious equations​

All about simulitious equations