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3 years ago

Necesito encontrar expresiones equivalentes a los valores del 1 al 20 pliss

Mathematics

1 answer:

Mariulka [41]3 years ago

4 0

Answer and Step-by-step explanation:

Si está buscando una expresión que sea igual a los números del 1 al 20, intente hacer una. Por ejemplo, 2y + 1x = 18 donde y = 9 - x / 2 y x = 18 - 2y. Siempre que la expresión sea igual a menos de 20, ¿no?

Lo siento, no entendí muy bien la pregunta, ¡pero espero que esto te haya ayudado! ¡Ten un día maravilloso! <3

**Tenga en cuenta que mi español no es el mejor, lo siento mucho, hablo principalmente inglés. Mi mal, con suerte, me puedes entender.**

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Find the equation of a line perpendicular to x=6+6y that passes through the point (-8,6)

alexgriva [62]

Answer:

y = -6x - 42

Step-by-step explanation:

First solve for y in x = 6 + 6y.

6y + 6 = x

6y = x - 6

6/6y = x/6 - 6/6

y = 1/6x - 1 the slope is 1/6.

Perpendicular lines slopes are negative reciprocals of each other.

So the new equation will have the slope of -6 ( the negative reciprocal of 1/6)

Next solve for "b" by using y = mx + b and the new slope of -6 and the point (-8,6)

y = mx + b

6 = -6(-8) + b

6 = 48 + b

-42 = b

The equation of the line is y = -6x - 42

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2 years ago

An extremely handsome math teacher was on a farm and he counted 32 animals; all pigs and chickens. Then he counted the number of

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Answer:

12 pigs and 20 chickens

Step-by-step explanation:

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Pig = 4 legs

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3 years ago

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Answer:

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3 years ago

Use the Newton-Raphson method to find the root of the equation f(x) = In(3x) + 5x2, using an initial guess of x = 0.5 and a stop

xxMikexx [17]

Answer with explanation:

The equation which we have to solve by Newton-Raphson Method is,

f(x)=log (3 x) +5 x²

$f'(x)=\frac{1}{3x}+10 x$

Initial Guess =0.5

Formula to find Iteration by Newton-Raphson method

$x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}\\\\x_{1}=x_{0}-\frac{f(x_{0})}{f'(x_{0})}\\\\ x_{1}=0.5-\frac{\log(1.5)+1.25}{\frac{1}{1.5}+10 \times 0.5}\\\\x_{1}=0.5- \frac{0.1760+1.25}{0.67+5}\\\\x_{1}=0.5-\frac{1.426}{5.67}\\\\x_{1}=0.5-0.25149\\\\x_{1}=0.248$

$x_{2}=0.248-\frac{\log(0.744)+0.30752}{\frac{1}{0.744}+10 \times 0.248}\\\\x_{2}=0.248- \frac{-0.128+0.30752}{1.35+2.48}\\\\x_{2}=0.248-\frac{0.17952}{3.83}\\\\x_{2}=0.248-0.0468\\\\x_{2}=0.2012$

$x_{3}=0.2012-\frac{\log(0.6036)+0.2024072}{\frac{1}{0.6036}+10 \times 0.2012}\\\\x_{3}=0.2012- \frac{-0.2192+0.2025}{1.6567+2.012}\\\\x_{3}=0.2012-\frac{-0.0167}{3.6687}\\\\x_{3}=0.2012+0.0045\\\\x_{3}=0.2057$

$x_{4}=0.2057-\frac{\log(0.6171)+0.21156}{\frac{1}{0.6171}+10 \times 0.2057}\\\\x_{4}=0.2057- \frac{-0.2096+0.21156}{1.6204+2.057}\\\\x_{4}=0.2057-\frac{0.0019}{3.6774}\\\\x_{4}=0.2057-0.0005\\\\x_{4}=0.2052$

So, root of the equation =0.205 (Approx)

Approximate relative error

$=\frac{\text{Actual value}}{\text{Given Value}}\\\\=\frac{0.205}{0.5}\\\\=0.41$

Approximate relative error in terms of Percentage

=0.41 × 100

= 41 %

7 0

3 years ago