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3 years ago

Slope form m=-3 (-2,1)

Mathematics

1 answer:

Hunter-Best [27]3 years ago

7 0

Answer:

y = -3x - 5

Step-by-step explanation:

y - 1 = -3 (x + 2)

y - 1 = -3x - 6

y = -3x -5

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Hey therapy was conducted to see how many people ride a bus to school every day. 160 people reported that they ride the bus to s

xeze [42]

Answer:

200 people

Step-by-step explanation:

Let: the number of people who surveyed be: x

Therefore, (160/x) x 100=80%

=> x=(160x100)/80

=>x=16000/80

=>x= 200

Thus, 200 people surveyed.

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3 years ago

True or false: is ST parallel to PR? (photo above)

Deffense [45]

Answer:

true

Step-by-step explanation:

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3 years ago

Read 2 more answers

What are the side lengths of the equilateral triangle?

Alchen [17]

$5x + (6x - 5) + (3x + 10) = 360$

$5x + 6x - 5 + 3x + 10 = 360$

$11x - 5 + 3x + 10 = 360$

$14x + 5 = 360$

$14x = 360 - 5$

$14x = 355$

$x = 25.36$

$x = 25.36$

$5x = 5 \times 25.36 = 126.8$

$6x - 5 = 6 \times 25.36 - 5 = 147.16$

$3x + 10 = 3 \times 25.36 + 10 = 86.08$

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3 years ago

How many nonzero terms of the Maclaurin series for ln(1 x) do you need to use to estimate ln(1.4) to within 0.001?

Vilka [71]

Answer:

The estimate of In(1.4) is the first five non-zero terms.

Step-by-step explanation:

From the given information:

We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

So, by the application of Maclurin Series which can be expressed as:

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2 f"(0)}{2!}+ \dfrac{x^3f'(0)}{3!}+... \ \ \ \ \ --- (1)$

Let examine f(x) = In(1+x), then find its derivatives;

f(x) = In(1+x)

$f'(x) = \dfrac{1}{1+x}$

f'(0) $= \dfrac{1}{1+0}=1$

f ' ' (x) $= \dfrac{1}{(1+x)^2}$

f ' ' (x) $= \dfrac{1}{(1+0)^2}=-1$

f ' ' '(x) $= \dfrac{2}{(1+x)^3}$

f ' ' '(x) $= \dfrac{2}{(1+0)^3} = 2$

f ' ' ' '(x) $= \dfrac{6}{(1+x)^4}$

f ' ' ' '(x) $= \dfrac{6}{(1+0)^4}=-6$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+x)^5} = 24$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+0)^5} = 24$

Now, the next process is to substitute the above values back into equation (1)

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2f' \ '(0)}{2!}+\dfrac{x^3f \ '\ '\ '(0)}{3!}+\dfrac{x^4f '\ '\ ' \ ' \(0)}{4!}+\dfrac{x^5f' \ ' \ ' \ ' \ '0)}{5!}+ ...$

$In(1+x) = o + \dfrac{x(1)}{1!}+ \dfrac{x^2(-1)}{2!}+ \dfrac{x^3(2)}{3!}+ \dfrac{x^4(-6)}{4!}+ \dfrac{x^5(24)}{5!}+ ...$

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

To estimate the value of In(1.4), let's replace x with 0.4

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

$In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...$

Therefore, from the above calculations, we will realize that the value of $\dfrac{0.4^5}{5}= 0.002048$ as well as $\dfrac{0.4^6}{6}= 0.00068267$ which are less than 0.001