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3 years ago

What is the slope of the line with equation y + 2 = 5 (x - 3) ?

Mathematics

1 answer:

Dahasolnce [82]3 years ago

3 0

Answer:

Step-by-step explanation:

y + 2 = 5x - 15

y = 5x - 17

slope is 5

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Solve the equation <br><br> 3x-1/2y=2

poizon [28]

Answer:

Step-by-step explanation:

Without a second equation relating x and y, we can solve 3x - 1/2y = 2 ONLY for x in terms of y or for y in terms of x:

x in terms of y: Multiply all three terms of 3x - 1/2y = 2 by 2, to eliminate the fraction: 6x - y = 4. Now add y to both sides to isolate 6x: 6x = 4 + y.

Last, divide both sides by 6 to isolate x:

x = (4 + y)/6

y in terms of x:

y = 6x - 4

If you want a numerical solution, please provide another equation in x and y and solve the resulting system.

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3 years ago

Given 9______

34kurt

7____ is the one that couldn't used to complete a table of eqiuvalent ratios

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3 years ago

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Point J is on line segment I K ‾ IK . Given J K = x + 6 , JK=x+6, I J = 9 , IJ=9, and I K = 2 x , IK=2x, determine the numerical

denpristay [2]

Answer:21

Step-by-step explanation:

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3 years ago

3(5j+2)=2(3j-6) how do i do this

ohaa [14]

3(5j+2)=2(3j-6) do distributed to remove parenthesis 15j+6=6j-3 . Then move all j to left side and the second term to right side 15j-6j=-3-6. Now solve both sides 9j= -9. Now divide both sides by 9 to get j alone to j = -1

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2 years ago

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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

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.

8 0

2 years ago