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2 years ago

Please help me with this.

Mathematics

1 answer:

inessss [21]2 years ago

8 0

Answer:

d. x = 1 1/3

Step-by-step explanation:

Step 1: Add 5/6x to both sides.

$\frac{2}{3}x + \frac{5}{6}x = 2 - \frac{5}{6}x + \frac{5}{6}x$
$\frac{3}{2}x = 2$

Step 2: Multiply both sides by 2/3.

$\frac{3}{2}x * \frac{2}{3} = 2 * \frac{2}{3}$
$x = \frac{4}{3}$
$x = 1 \frac{1}{3}$

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If z=3 and z*y=12 what does y+5 equal

mart [117]

Z = 3 (This equation is a given)

z(y) = 12 ( This equation is also given)

3y = 12 (as z = 3, plug 3 into z)

3y/3 = 12/3 (isolate the y by dividing 3 to both sides)
* you divide 3 to both sides because of the equal sign. What you do to one side you do to the other
y = 4 (answer for y)

y + 5 (what you are solving for)

(4) + 5 = 9 (As y = 4, plug in 4 for y in this equation)

y + 5 = 9 (4 + 5 = 9, or addition property)

9 is your answer

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Step-by-step explanation:

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Find the expansion of tan x about the point X = 0.

yan [13]

Answer:

$f(x) = x +\frac{1}{3}x^{3}+....$

Step-by-step explanation:

As per the question,

let us consider f(x) = tan(x).

We know that <u>The Maclaurin series is given by:</u>

$f(x) = f(0) + \frac{f^{'}(0)}{1!}\cdot x+ \frac{f^{''}(0)}{2!}\cdot x^{2}+\frac{f^{'''}(0)}{3!}\cdot x^{3}+......$

So, differentiate the given function 3 times in order to find f'(x), f''(x) and f'''(x).

Therefore,

f'(x) = sec²x

f''(x) = 2 × sec(x) × sec(x)tan(x)

= 2 × sec²(x) × tan(x)

f'''(x) = 2 × 2 sec²(x) tan(x) tan(x) + 2 sec²(x) × sec²(x)

= 4sec²(x) tan²(x) + 2sec⁴(x)

= 6 sec⁴x - 4 sec² x

We then substitute x with 0, and find the values

f(0) = tan 0 = 0

f'(0) = sec²0 = 1

f''(0) = 2 × sec²(0) × tan(0) = 0

f'''(0) = 6 sec⁴0- 4 sec² 0 = 2

By putting all the values in the Maclaurin series, we get

$f(x) = f(0) + \frac{f^{'}(0)}{1!}\cdot x+ \frac{f^{''}(0)}{2!}\cdot x^{2}+\frac{f^{'''}(0)}{3!}\cdot x^{3}+......$

$f(x) = 0 + \frac{1}{1}\cdot x+ \frac{0}{2}\cdot x^{2}+\frac{2}{6}\cdot x^{3}+......$

$f(x) = x +\frac{1}{3}x^{3}+....$

Therefore, the expansion of tan x at x = 0 is

$f(x) = x +\frac{1}{3}x^{3}+....$ .

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