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2 years ago

When solving negative one over three (x − 15) = −4, what is the correct sequence of operations?

1 answer:

iogann1982 [59]2 years ago

7 0

-1/3(x - 15) = -4

step 1 : multiply both sides by -3, step 2 : add 15 to both sides

because when u multiply both sides by -3, this cancels out the -1/3 on the left side leaving u with : x - 15 = -4 * -3.....x - 15 = 12.....and then u would add 15 to both sides....giving u : x = 12 + 15.....x = 27

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nikdorinn [45]

9514 1404 393

Answer:

maximum difference is 38 at x = -3

Step-by-step explanation:

This is nicely solved by a graphing calculator, which can plot the difference between the functions. The attached shows the maximum difference on the given interval is 38 at x = -3.

Ordinarily, the distance between curves is measured vertically. Here that means you're interested in finding the stationary points of the difference between the functions, along with that difference at the ends of the interval. The maximum difference magnitude is what you're interested in.

h(x) = g(x) -f(x) = (2x³ +5x² -15x) -(x³ +3x² -2) = x³ +2x² -15x +2

Then the derivative is ...

h'(x) = 3x² +4x -15 = (x +3)(3x -5)

This has zeros (stationary points) at x = -3 and x = 5/3. The values of h(x) of concern are those at x=-5, -3, 5/3, 3. These are shown in the attached table.

The maximum difference between f(x) and g(x) is 38 at x = -3.

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Answer:

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2 years ago

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EleoNora [17]

Answer:

Step-by-step explanation:

hello

$a_1=18\\\\a_2=2+\dfrac{a_1}{2}=2+\dfrac{18}{2}=2+9=11\\\\a_3=2+\dfrac{a_2}{2}=2+\dfrac{11}{2}=\dfrac{4+11}{2}=\dfrac{15}{2}\\\\a_4=2+\dfrac{a_3}{2}=2+\dfrac{15}{2}=\dfrac{4+15}{2}=\dfrac{19}{2}$

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2 years ago

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