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2 years ago

Please, someone. Solve for x.

Mathematics

1 answer:

grin007 [14]2 years ago

7 0

Combine the fractions on the left side:

9/(x + 2) + 2/(x - 2) = 1

9 (x - 2) / ((x + 2) (x - 2)) + 2 (x + 2) / ((x + 2) (x - 2)) = 1

(9x - 18) / ((x + 2) (x - 2)) + (2x + 4) / ((x + 2) (x - 2)) = 1

(9x - 18 + 2x + 4) / ((x + 2) (x - 2)) = 1

(11x - 14) / ((x + 2) (x - 2)) = 1

Recall the difference of squares identity,

a² - b² = (a + b) (a - b)

which lets us simplify the denominator on the left side as

(11x - 14) / (x² - 4) = 1

For any real numbers a and b, if a/b = 1, then a = b. This means

11x - 14 = x² - 4

which we can rearrange as

x² - 11x + 10 = 0

Factorize the left side:

(x - 10) (x - 1) = 0

Then

x - 10 = 0 or x - 1 = 0

x = 10 or x = 1

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A spring has natural length 0.75 m and a 5 kg mass. A force of 8 N is needed to keep the spring stretched to a length of 0.85 m.

I am Lyosha [343]

Use Hooke's law to find the spring constant. If it takes 8N to stretch the spring by 0.1m, then

$8\text{ N}=k(0.1\text{ m})\implies k=80\dfrac{\text N}{\text m}$

I'm going to assume the spring is fixed to a ceiling, and that any stretching in the downward direction counts as movement in the positive direction. The spring's motion is then modeled by

$y''(t)+80y(t)=0$

where $y(t)$ is the position of the spring's free end as it moves up and down. Solving this is easy enough: the characteristic solution will be

$y(t)=C_1\cos4\sqrt5t+C_2\sin4\sqrt5t$

Given that the spring is stretched to a length of 1m (a difference of 0.25m from its natural length), and is released with no external pushing or pulling, we have the two initial conditions $y(0)=\dfrac14$ and $y'(0)=0$ .

$y(0)=\dfrac14\implies\dfrac14=C_1\cos0+C_2\sin0\implies C_1=\dfrac14$
$y'(0)=0\implies 0=-4\sqrt5C_1\sin0+4\sqrt5C_2\cos0\implies C_2=0$

So the spring's motion is dictated by the function

$y(t)=\dfrac14\cos4\sqrt5t$

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3 years ago

The line with equation −a+3b=0 coincides with the terminal side of an angle θ in standard position and sinθ<0 .

Anna11 [10]

If $a$ is the variable of the horizontal axis, then you can solve for $b$ to get the equation of the line in slope-intercept form in the $a,b$ plane:

$-a+3b=0\implies b=\dfrac a3$

i.e. a line with slope $\dfrac13$ through the origin, which means it is contained in the first and third quadrants. Since the terminal side of $\theta$ has a negative sine, the angle must lie in the third quadrant.

Because the slope of the line is $\dfrac13$ , you can choose any length along the line to make up the hypotenuse of a right triangle with reference angle $\theta$ . Any such right triangle will have $\tan\theta=\dfrac13$ , regardless of whether the angle is the first or third quadrant. But since $\theta$ is known to lie in the third quadrant, and so $\sin\theta$ and $\cos\theta$ are both negative, you have

$\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac13\implies \dfrac{\sin\theta}{\cos\theta}=\dfrac{-1}{-3}\implies \cos\theta=-3$

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4 years ago

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sergejj [24]

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$\frac{13-5}{6-2}$
$\frac{8}{4}
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So the slope of the line that passes through that point is 2.

Hope this helps :)

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4 years ago

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Please 50 points and brainliest youre rank will go up

EastWind [94]

Answer:

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Step-by-step explanation:

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3 years ago

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Taya2010 [7]

Answer:

Option C. failing to reject a false null hypothesis

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Type II error states that Probability of accepting null hypothesis given the fact that null hypothesis is false.

This is considered to be the most important error.

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3 years ago