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3 years ago

Solve the following equation for x. x^2 - 36 = 0

Mathematics

1 answer:

Ilia_Sergeevich [38]3 years ago

4 0

X = -6, x = 6
option c

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3x+6/x^2-x-6+2x/x^2+x-12

Tanzania [10]

$\dfrac{3x+6}{x^2-x-6} + \dfrac{2x}{x^2+x-12}$

$= \dfrac{3(x+2)}{(x + 2)(x - 3)} + \dfrac{2x}{ (x - 3)(x + 4)}$

$= \dfrac{3}{(x - 3)} + \dfrac{2x}{ (x -3)(x + 4)}$

$= \dfrac{3(x+4)}{(x - 3)(x + 4)} + \dfrac{2x}{ (x -3 )(x + 4)}$

$= \dfrac{3(x+4) + 2x }{(x - 3)(x + 4)}$

$= \dfrac{3x+12 + 2x }{(x - 3)(x + 4)}$

$= \dfrac{5x+12}{(x - 3)(x + 4)}$

6 0

3 years ago

Cual es el 15% de 36000?

zimovet [89]

15% = 15/100 = 0,15

36000•0,15=5400

4 0

3 years ago

Find the distance between the points given. (0,5) and (-5,0)

galina1969 [7]

Answer:

5 $\sqrt{2}$

Step-by-step explanation:

Calculate the distance between the points using the distance formula

d = √ (x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = (0, 5) and (x₂, y₂ ) = (- 5, 0)

d = $\sqrt{(-5-0)^2+(0-5)^2}$

= $\sqrt{(-5)^2+(-5)^2}$

= $\sqrt{25+25}$ = $\sqrt{50}$ = 5 $\sqrt{2}$ ← exact value

5 0

3 years ago

How to solve this trig

n200080 [17]

Hi there!

To find the Trigonometric Equation, we have to isolate sin, cos, tan, etc. We are also given the interval [0,2π).

First Question

What we have to do is to isolate cos first.

$\displaystyle \large{ cos \theta = - \frac{1}{2} }$

Then find the reference angle. As we know cos(π/3) equals 1/2. Therefore π/3 is our reference angle.

Since we know that cos is negative in Q2 and Q3. We will be using π + (ref. angle) for Q3. and π - (ref. angle) for Q2.

Find Q2

$\displaystyle \large{ \pi - \frac{ \pi}{3} = \frac{3 \pi}{3} - \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{2 \pi}{3} }$

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{3} = \frac{3 \pi}{3} + \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{4 \pi}{3} }$

Both values are apart of the interval. Hence,

$\displaystyle \large \boxed{ \theta = \frac{2 \pi}{3} , \frac{4 \pi}{3} }$

Second Question

Isolate sin(4 theta).

$\displaystyle \large{sin 4 \theta = - \frac{1}{ \sqrt{2} } }$

Rationalize the denominator.

$\displaystyle \large{sin4 \theta = - \frac{ \sqrt{2} }{2} }$

The problem here is 4 beside theta. What we are going to do is to expand the interval.

$\displaystyle \large{0 \leqslant \theta < 2 \pi}$

Multiply whole by 4.

$\displaystyle \large{0 \times 4 \leqslant \theta \times 4 < 2 \pi \times 4} \\ \displaystyle \large \boxed{0 \leqslant 4 \theta < 8 \pi}$

Then find the reference angle.

We know that sin(π/4) = √2/2. Hence π/4 is our reference angle.

sin is negative in Q3 and Q4. We use π + (ref. angle) for Q3 and 2π - (ref. angle for Q4.)

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{4} = \frac{ 4 \pi}{4} + \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{5 \pi}{4} }$

Find Q4

$\displaystyle \large{2 \pi - \frac{ \pi}{4} = \frac{8 \pi}{4} - \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{7 \pi}{4} }$

Both values are in [0,2π). However, we exceed our interval to < 8π.

We will be using these following:-

$\displaystyle \large{ \theta + 2 \pi k = \theta \: \: \: \: \: \sf{(k \: \: is \: \: integer)}}$

Hence:-

For Q3

$\displaystyle \large{ \frac{5 \pi}{4} + 2 \pi = \frac{13 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 4\pi = \frac{21 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 6\pi = \frac{29 \pi}{4} }$

We cannot use any further k-values (or k cannot be 4 or higher) because it'd be +8π and not in the interval.

For Q4

$\displaystyle \large{ \frac{ 7 \pi}{4} + 2 \pi = \frac{15 \pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 4 \pi = \frac{23\pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 6 \pi = \frac{31 \pi}{4} }$

Therefore:-

$\displaystyle \large{4 \theta = \frac{5 \pi}{4} , \frac{7 \pi}{4} , \frac{13\pi}{4} , \frac{21\pi}{4} , \frac{29\pi}{4}, \frac{15 \pi}{4} , \frac{23\pi}{4} , \frac{31\pi}{4} }$

Then we divide all these values by 4.

$\displaystyle \large \boxed{\theta = \frac{5 \pi}{16} , \frac{7 \pi}{16} , \frac{13\pi}{16} , \frac{21\pi}{16} , \frac{29\pi}{16}, \frac{15 \pi}{16} , \frac{23\pi}{16} , \frac{31\pi}{16} }$

Let me know if you have any questions!

3 0

3 years ago

What is the sum of the first five terms in the series 7+12+17+..?

nirvana33 [79]

7 + 12 + 17 + 22 + 27 = ( 7 + 27 ) + ( 12 + 22 ) + ( 34 / 2 ) = ( 34 / 1 ) + ( 34 / 1 ) + ( 34 / 2 ) = ( 68 / 2 ) + ( 68 / 2 ) + ( 34 / 2 ) = 170 / 2 = 85 .

3 0

3 years ago