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3 years ago

Does anyone know how to do this ??? -view attachment

Mathematics

1 answer:

cricket20 [7]3 years ago

8 0

As soon as I read this, the words "law of cosines" popped
into my head. I don't have a good intuitive feeling for the
law of cosines, but I went and looked it up (you probably
could have done that), and I found that it's exactly what
you need for this problem.

The "law of cosines" relates the lengths of the sides of any
triangle to the cosine of one of its angles ... just what we need,
since we know all the sides, and we want to find one of the angles.

To find angle-B, the law of cosines says

   b² = a² + c² - 2 a c cosine(B)

B = angle-B
b = the side opposite angle-B = 1.4
a, c = the other 2 sides = 1 and 1.9

(1.4)² = (1)² + (1.9)² - (2 x 1 x 1.9) cos(B)

   1.96 = (1) + (3.61) - (3.8) cos(B)

Add 3.8 cos(B) from each side:

   1.96 + 3.8 cos(B) = 4.61

Subtract 1.96 from each side:

   3.8 cos(B) = 2.65

Divide each side by 3.8 :

cos(B) = 0.69737 (rounded)

Whipping out the
trusty calculator:
   B = the angle whose cosine is 0.69737

      = 45.784° .

Now, for the first time, I'll take a deep breath, then hold it
while I look back at the question and see whether this is
anywhere near one of the choices ...

By gosh ! Choice 'B' is 45.8° ! yay !
I'll bet that's it !

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Please help right away.

Travka [436]

$\sum\limits_{k=1}^{\infty}420\left(\dfrac{1}{6}\right)^{k-1}$

The infinite geometric series is converges if |r| < 1.

We have r=1/6 < 1, therefore our infinite geometric series is converges.

The sum S of an infinite geometric series with |r| < 1 is given by the formula :

$S=\dfrac{a_1}{1-r}$

We have:

$a_1=420\left(\dfrac{1}{6}\right)^{1-1}=420\left(\dfrac{1}{6}\right)^0=420\\\\r=\dfrac{1}{6}$

substitute:

$S=\dfrac{420}{1-\frac{1}{6}}=\dfrac{420}{\frac{5}{6}}=420\cdot\dfrac{6}{5}=84\cdot6=504$

Answer: d. Converges, 504.

8 0

3 years ago

Que is on pic.i can't able to type in text.

ad-work [718]

It's not difficult to compute the values of $A$ and $B$ directly:

$A=\displaystyle\int_1^{\sin\theta}\frac{\mathrm dt}{1+t^2}=\tan^{-1}t\bigg|_{t=1}^{t=\sin\theta}$
$A=\tan^{-1}(\sin\theta)-\dfrac\pi4$

$B=\displaystyle\int_1^{\csc\theta}\frac{\mathrm dt}{t(1+t^2)}=\int_1^{\csc\theta}\left(\frac1t-\frac t{1+t^2}\right)\,\mathrm dt$
$B=\left(\ln|t|-\dfrac12\ln|1+t^2|\right)\bigg|_{t=1}^{t=\csc\theta}$
$B=\ln\left|\dfrac{\csc\theta}{\sqrt{1+\csc^2\theta}}\right|+\dfrac12\ln2$

Let's assume $0$ , so that $|\csc\theta|=\csc\theta$ .

Now,

$\Delta=\begin{vmatrix}A&A^2&B\\e^{A+B}&B^2&-1\\1&A^2+B^2&-1\end{vmatrix}$
$\Delta=A\begin{vmatrix}B^2&-1\\A^2+B^2&-1\end{vmatrix}-e^{A+B}\begin{vmatrix}A^2&B\\A^2+B^2&-1\end{vmatrix}+\begin{vmatrix}A^2&B\\B^2&-1\end{vmatrix}$
$\Delta=A(-B^2+A^2+B^2)-e^{A+B}(-A^2-A^2B-B^3)+(-A^2-B^3)$
$\Delta=A^3-A^2-B^3+e^{A+B}(A^2+A^2B+B^3)$

There doesn't seem to be anything interesting about this result... But all that's left to do is plug in $A$ and $B$ .

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Find the greatest common factor for 10,40

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Travka [436]

Answer:

The answer is 0, 1/3.

Step-by-step explanation:

This is because only at 0 does the line intersect through 1/3.

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Answer:

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Step-by-step explanation:

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