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3 years ago

Find measure dbc 19 22 23 161 HELP !!!

Mathematics

1 answer:

uysha [10]3 years ago

4 0

∠DBC = 19° since the two angles have to add up to 180, if x = 23, then 23 -4 = 19 for the right, and 7 x 23 = 161, and 161 + 19 = 180.

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\lim _{x\to 0}\left(\frac{2x\ln \left(1+3x\right)+\sin \left(x\right)\tan \left(3x\right)-2x^3}{1-\cos \left(3x\right)}\right)

Vinvika [58]

$\displaystyle \lim_{x\to 0}\left(\frac{2x\ln \left(1+3x\right)+\sin \left(x\right)\tan \left(3x\right)-2x^3}{1-\cos \left(3x\right)}\right)$

Both the numerator and denominator approach 0, so this is a candidate for applying L'Hopital's rule. Doing so gives

$\displaystyle \lim_{x\to 0}\left(2\ln(1+3x)+\dfrac{6x}{1+3x}+\cos(x)\tan(3x)+3\sin(x)\sec^2(x)-6x^2}{3\sin(3x)}\right)$

This again gives an indeterminate form 0/0, but no need to use L'Hopital's rule again just yet. Split up the limit as

$\displaystyle \lim_{x\to0}\frac{2\ln(1+3x)}{3\sin(3x)} + \lim_{x\to0}\frac{6x}{3(1+3x)\sin(3x)} \\\\ + \lim_{x\to0}\frac{\cos(x)\tan(3x)}{3\sin(3x)} + \lim_{x\to0}\frac{3\sin(x)\sec^2(x)}{3\sin(3x)} \\\\ - \lim_{x\to0}\frac{6x^2}{3\sin(3x)}$

Now recall two well-known limits:

$\displaystyle \lim_{x\to0}\frac{\sin(ax)}{ax}=1\text{ if }a\neq0 \\\\ \lim_{x\to0}\frac{\ln(1+ax)}{ax}=1\text{ if }a\neq0$

Compute each remaining limit:

$\displaystyle \lim_{x\to0}\frac{2\ln(1+3x)}{3\sin(3x)} = \frac23 \times \lim_{x\to0}\frac{\ln(1+3x)}{3x} \times \lim_{x\to0}\frac{3x}{\sin(3x)} = \frac23$

$\displaystyle \lim_{x\to0}\frac{6x}{3(1+3x)\sin(3x)} = \frac23 \times \lim_{x\to0}\frac{3x}{\sin(3x)} \times \lim_{x\to0}\frac{1}{1+3x} = \frac23$

$\displaystyle \lim_{x\to0}\frac{\cos(x)\tan(3x)}{3\sin(3x)} = \frac13 \times \lim_{x\to0}\frac{\cos(x)}{\cos(3x)} = \frac13$

$\displaystyle \lim_{x\to0}\frac{3\sin(x)\sec^2(x)}{3\sin(3x)} = \frac13 \times \lim_{x\to0}\frac{\sin(x)}x \times \lim_{x\to0}\frac{3x}{\sin(3x)} \times \lim_{x\to0}\sec^2(x) = \frac13$

$\displaystyle \lim_{x\to0}\frac{6x^2}{3\sin(3x)} = \frac23 \times \lim_{x\to0}x \times \lim_{x\to0}\frac{3x}{\sin(3x)} \times \lim_{x\to0}x = 0$

So, the original limit has a value of

2/3 + 2/3 + 1/3 + 1/3 - 0 = 2

6 0

3 years ago

I need help with this logic puzzle,

natka813 [3]

I cannot reach a meaningful solution from the given information. To prove that S was always true, you would have to prove that N was always false. To prove that N was always false you would have to prove that L was always false. For the statement (L ^ T) -> K to be true, you only need K to be true, so L can be either true or false.
Therefore, because of the aforementioned knowledge, I do not believe that you can prove S to be true.

3 0

4 years ago

Given y=(2x+3)2, choose the standard form of the given quadratic equation

katrin2010 [14]

$y=(2x+3)^2 = 4x^2 +12x +9$

Taking 4 out from first two terms, we will get

$y=4(x^2+3x)+9$

Now we divide the coefficient of x by 2 and add and subtract the square of the result, we will get

$y=4(x^2+3x+\frac{9}{4}-\frac{9}{4})+9$

Distributing 4

$y=4(x+\frac{3}{2})^2+9-9$

Cancelling 9

$y=4(x+\frac{3}{2})^2$

And that's the required standard form.

5 0

3 years ago

Number 14 on my integrated geometry test

FinnZ [79.3K]

Answer:

C. 21

Step-by-step explanation:

$\cos \: 30 \degree = \frac{x}{24} \\ \\ \frac{ \sqrt{3} }{2} = \frac{x}{24} \\ \\ x = \frac{24 \sqrt{3} }{2} \\ \\ x = 12 \sqrt{3} \\ \\ x = 20.7846097 \\ \\ x \approx \: 21$