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3 years ago

A line with a slope of three is written in the form y=Mx+b what is the value of be if the line passes through the point (3,11)

Mathematics

2 answers:

lana [24]3 years ago

6 0

Answer:

y=3x+2

Step-by-step explanation:

y-y1=m(x-x1)

y-11=3(x-3)

y=3x-9+11

y=3x+2

zepelin [54]3 years ago

4 0

Answer: y=3x+b

Let’s rewrite the information we have

Model equation: y=mx+b

Slope: 3

Point: (3,11)

In the model equation, this is what the letters stand for— y=slope•variable+y-intercept. Since we already know the slope, let’s substitute this into the equation now.

y=3x+b

Now we need to find the y-intercept/b. We’ll do this by substituting the y and x value from the point into the equation and solving for b

y=3x+b

*substitute point*

11=3(3)+b

*multiply 3 and 3*

11=9+b

*subtract 9 on both sides

2=b

Now that we have b, let’s put this into the equation and finish the equation

y=3x+b

Hope this helps comment below for more questions :)

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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;

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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)

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

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

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

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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

Hence, the estimate of In(1.4) to the term is $\dfrac{0.4^5}{5}$ is said to be enough to justify our claim.

∴

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

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